M53 E3 Reviewer.pdf

Unit III Exam Reviewer

Mathematics 53

Using the Mean Value Theorem, show that for the function

f(x) = x

2 _+4x

x−7 , there exists a c in the interval (2, 6) for

which f0(c) = −72

5 .

Isf(x)continuous in [2, 6]? YES

Isf(x)differentiable in(2, 6)? YES

f0(x) = (x−7)(2x+4)−(x

2_+4x)(1)

(x−7)2 =

x2−14x−28 (x−7)2

By the MVT, there exists a number c∈(2, 6)such that

f0(c) = f(6)−f(2)

6−2 =

60

−1− −125

4 =

−288 5

4 = −

Using the Mean Value Theorem, show that for the function

f(x) = x

2 _+4x

x−7 , there exists a c in the interval (2, 6) for

which f0(c) = −72

5 .

Isf(x)continuous in [2, 6]?

YES

Isf(x)differentiable in(2, 6)? YES

f0(x) = (x−7)(2x+4)−(x

2_+4x)(1)

(x−7)2 =

x2−14x−28 (x−7)2

By the MVT, there exists a number c∈(2, 6)such that

f0(c) = f(6)−f(2)

6−2 =

60

−1− −125

4 =

−288 5

4 = −

Using the Mean Value Theorem, show that for the function

f(x) = x

2 _+4x

x−7 , there exists a c in the interval (2, 6) for

which f0(c) = −72

5 .

Isf(x)continuous in [2, 6]? YES

Isf(x)differentiable in(2, 6)? YES

f0(x) = (x−7)(2x+4)−(x

2_+4x)(1)

(x−7)2 =

x2−14x−28 (x−7)2

By the MVT, there exists a number c∈(2, 6)such that

f0(c) = f(6)−f(2)

6−2 =

60

−1− −125

4 =

−288 5

4 = −

Using the Mean Value Theorem, show that for the function

f(x) = x

2 _+4x

x−7 , there exists a c in the interval (2, 6) for

which f0(c) = −72

5 .

Isf(x)continuous in [2, 6]? YES

Isf(x)differentiable in(2, 6)?

YES

f0(x) = (x−7)(2x+4)−(x

2_+4x)(1)

(x−7)2 =

x2−14x−28 (x−7)2

By the MVT, there exists a number c∈(2, 6)such that

f0(c) = f(6)−f(2)

6−2 =

60

−1− −125

4 =

−288 5

4 = −

Using the Mean Value Theorem, show that for the function

f(x) = x

2 _+4x

x−7 , there exists a c in the interval (2, 6) for

which f0(c) = −72

5 .

Isf(x)continuous in [2, 6]? YES

Isf(x)differentiable in(2, 6)? YES

f0(x) = (x−7)(2x+4)−(x

2_+4x)(1)

(x−7)2 =

x2_−14x−28 (x−7)2

By the MVT, there exists a number c∈(2, 6)such that

f0(c) = f(6)−f(2)

6−2 =

60

−1− −125

4 =

−288 5

4 = −

Using the Mean Value Theorem, show that for the function

f(x) = x

2 _+4x

x−7 , there exists a c in the interval (2, 6) for

which f0(c) = −72

5 .

Isf(x)continuous in [2, 6]? YES

Isf(x)differentiable in(2, 6)? YES

f0(x) = (x−7)(2x+4)−(x

2_+4x)(1)

(x−7)2 =

x2_−14x−28 (x−7)2

By the MVT, there exists a number c∈(2, 6) such that

f0(c) = f(6)−f(2)

6−2 =

60

−1− −125

4 =

−288 5

4 = −

Using the Mean Value Theorem, show that for the function

f(x) = x

2 _+4x

x−7 , there exists a c in the interval (2, 6) for

which f0(c) = −72

5 .

Isf(x)continuous in [2, 6]? YES

Isf(x)differentiable in(2, 6)? YES

f0(x) = (x−7)(2x+4)−(x

2_+4x)(1)

(x−7)2 =

x2_−14x−28 (x−7)2

By the MVT, there exists a number c∈(2, 6) such that

f0(c) = f(6)−f(2)

6−2 =

60

−1− −125

4

= −

288 5

4 = −

Using the Mean Value Theorem, show that for the function

f(x) = x

2 _+4x

x−7 , there exists a c in the interval (2, 6) for

which f0(c) = −72

5 .

Isf(x)continuous in [2, 6]? YES

Isf(x)differentiable in(2, 6)? YES

f0(x) = (x−7)(2x+4)−(x

2_+4x)(1)

(x−7)2 =

x2_−14x−28 (x−7)2

By the MVT, there exists a number c∈(2, 6) such that

f0(c) = f(6)−f(2)

6−2 =

60

−1− −125

4 =

−288 5

4 =−

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.

NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.

NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.

NONE

lim x→±∞

x3−4

x2₋₂

= lim

x→±∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.

NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.

NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.

NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂

=−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞

lim x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂

=∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.

x=√2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.x= √2

andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.x= √2

andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂

=∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.x= √2

andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞

lim x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.x= √2

andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

x2₋₂

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.x= √2

andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

H.A.NONE

lim x→±∞

x3−4

x2₋₂ =_x→±lim_∞

x3−4

x2₋₂ · 1

x2

1

x2

= lim

x→±∞

x−4

x 1− 2

x2

=±∞

V.A.x= √2 andx=−√2

lim x→√2+

x3−4

x2₋₂ =−∞ lim

x→√2−

x3−4

x2₋₂ =∞

lim x→−√2+

x3−4

x2₋₂ =∞ lim

x→−√2−

x3−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

O.A.

y=x

x x2₋₂

x3 ₋₄

−x3+2x

2x

x3−4

x2₋₂ =x+

2x−4

x2₋₂

lim x→∞

2x−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

O.A.

y=x

x x2₋₂

x3 ₋₄

−x3+2x

2x

x3−4

x2₋₂ =x+

2x−4

x2₋₂

lim x→∞

2x−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

O.A.

y=x

x x2₋₂

x3 ₋₄

−x3+2x

2x

x3−4

x2₋₂ =x+

2x−4

x2₋₂

lim x→∞

2x−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

O.A.

y=x

x x2₋₂

x3 ₋₄

−x3+2x

2x

x3−4

x2₋₂ =x+

2x−4

x2₋₂

lim x→∞

2x−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

O.A.

y=x

x x2₋₂

x3 ₋₄

−x3+2x

2x

x3−4

x2₋₂ =x+

2x−4

x2₋₂

lim x→∞

2x−4

x2₋₂

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

O.A.

y=x

x x2₋₂

x3 ₋₄

−x3+2x

2x

x3−4

x2₋₂ =x+

2x−4

x2₋₂

lim x→∞

2x−4

Find, if any, the vertical, horizontal and oblique asymptotes of f(x) = x

3₋₄

x2₋₂.

O.A.y=x

x x2₋₂

x3 ₋₄

−x3+2x

2x

x3−4

x2₋₂ =x+

2x−4

x2₋₂

lim x→∞

2x−4

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Is f continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.

x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]?

YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.

x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.

x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.

x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.

x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.

x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx

=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.

x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0

=⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.

x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and 5π 4 ,−

√

Determine, if any, the absolute extrema in the interval

[−2π, 2π] of f(x) = sinx+cosx.

Isf continuous in [−2π, 2π]? YES

By the Extreme Value Theorem (EVT), there exists an absolute maximum and an absolute minimum in [−2π, 2π].

C.N.x=−7π

4 ,− 3π 4 , π 4, 5π 4

f0(x) =cosx−sinx=0 =⇒ cosx=sinx

f

−7π

4

=fπ 4

=√2 f

₋ 3π 4 =f 5π 4

=−√2 f(±2π) =1

f has abs. maxima at−7π 4 ,

√

2andπ 4,

√

2, and abs. minima at

−3π 4 ,−

√

2and5π 4 ,−

√

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain

R− {−1, 4}

x-intercepts

x=1, 2

y-intercepts

y=−1 2

horizontal asymptote

y=1 lim

x→±∞f(x) =1

vertical asymptote

x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain

R− {−1, 4}

x-intercepts

x=1, 2

y-intercepts

y=−1 2

horizontal asymptote

y=1 lim

x→±∞f(x) =1

vertical asymptote

x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts

x=1, 2

y-intercepts

y=−1 2

horizontal asymptote

y=1 lim

x→±∞f(x) =1

vertical asymptote

x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts

y=−1 2

horizontal asymptote

y=1 lim

x→±∞f(x) =1

vertical asymptote

x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2 horizontal asymptote

y=1 lim

x→±∞f(x) =1

vertical asymptote

x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2 horizontal asymptote

y=1

lim

x→±∞f(x) =1

vertical asymptote

x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

horizontal asymptote y=1 lim

x→±∞f(x) =1

vertical asymptote

x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

horizontal asymptote y=1 lim

x→±∞f(x) =1

vertical asymptote

x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞

lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

horizontal asymptote y=1 lim

x→±∞f(x) =1

vertical asymptote x=−1

andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞

lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

horizontal asymptote y=1 lim

x→±∞f(x) =1

vertical asymptote x=−1

andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

horizontal asymptote y=1 lim

x→±∞f(x) =1

vertical asymptote x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

oblique asymptote

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

horizontal asymptote y=1 lim

x→±∞f(x) =1

vertical asymptote x=−1 andx=4

lim

x→−1−f(x) =∞ _x→−lim₁+f(x) =−∞ lim

x→4−f(x) =−∞ _xlim_→4+f(x) =∞

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

H.A. y=1

V.A. x=−1andx=4

O.A. NONE

f0(x) =− 6(2x−3)

(x+1)2_(x−4)2 f

00_{(x) =} 12(13−9x+3x2)

(−4+x)3_(1+x)3

CNs x= 3

2

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

H.A. y=1

V.A. x=−1andx=4

O.A. NONE

f0(x) =− 6(2x−3)

(x+1)2_(x−4)2 f

00_{(x) =} 12(13−9x+3x2)

(−4+x)3_(1+x)3

CNs x= 3

2

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

domain R− {−1, 4}

x-intercepts x=1, 2

y-intercepts y= −1 2

H.A. y=1

V.A. x=−1andx=4

O.A. NONE

f0(x) =− 6(2x−3)

(x+1)2_(x−4)2 f

00_{(x) =} 12(13−9x+3x2)

(−4+x)3_(1+x)3

CNs x= 3

2

Sketch the graph of f(x) = (x−2)(x−1) (x+1)(x−4).

Interval f(x) f0(x) f00(x) Remarks

(−∞,−1) + + inc cu

x=−1 und und und VA

−1,3₂

+ - inc cd

x= 3_{2 25}1 0 - rel. max.

3 2, 4

- - dec cd

x=4 und und und VA

(4,∞) - + dec cu

Note:

lim

x→−1−f(x) = ∞

lim

x→−1+f(x) = −∞ lim

x→4−f(x) = −∞

lim

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Let landhbe the length and height of the poster, respectively.

Objective: Minimize Aposter=lh

Constraints: Aprint=32= l−8₃

(h−4) =⇒ l= 32

h−4+ 8 3

Min A(h) =

32h h−4 +

8h

3

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Let landhbe the length and height of the poster, respectively.

Objective: Minimize Aposter=lh

Constraints: Aprint=32= l−8₃

(h−4) =⇒ l= 32

h−4+ 8 3

Min A(h) =

32h h−4 +

8h

3

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Let landhbe the length and height of the poster, respectively.

Objective:

MinimizeAposter=lh

Constraints: Aprint=32= l−8₃

(h−4) =⇒ l= 32

h−4+ 8 3

Min A(h) =

32h h−4 +

8h

3

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Let landhbe the length and height of the poster, respectively.

Objective: Minimize Aposter=lh

Constraints: Aprint=32= l−8₃

(h−4) =⇒ l= 32

h−4+ 8 3

Min A(h) =

32h h−4 +

8h

3

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Let landhbe the length and height of the poster, respectively.

Objective: Minimize Aposter=lh

Constraints:

Aprint=32= l−8₃

(h−4) =⇒ l= 32

h−4+ 8 3

Min A(h) =

32h h−4 +

8h

3

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Let landhbe the length and height of the poster, respectively.

Objective: Minimize Aposter=lh

Constraints: Aprint=32= l−8₃

(h−4)

=⇒ l= 32

h−4+ 8 3

Min A(h) =

32h h−4 +

8h

3

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Let landhbe the length and height of the poster, respectively.

Objective: Minimize Aposter=lh

Constraints: Aprint=32= l−8₃

(h−4) =⇒ l= 32

h−4+ 8 3

Min A(h) =

32h h−4 +

8h

3

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Let landhbe the length and height of the poster, respectively.

Objective: Minimize Aposter=lh

Constraints: Aprint=32= l−8₃

(h−4) =⇒ l= 32

h−4+ 8 3

Min A(h) =

32h h−4 +

8h

3

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Min A(h) =

32h h−4 +

8h

3

where h∈ (4,∞)

A0(h) = 8(h

2_−8h−32)

3(h−4)2 A

00_{(h) =} 256

(h−4)3

CNs: h=4+4√3 (h=4−4√3<0)

A00(4+4√3) = 256 (4√3)3 >0

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Min A(h) =

32h h−4 +

8h

3

where h∈ (4,∞)

A0(h) = 8(h

2_−8h−32)

3(h−4)2

A00(h) = 256 (h−4)3

CNs: h=4+4√3 (h=4−4√3<0)

A00(4+4√3) = 256 (4√3)3 >0

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Min A(h) =

32h h−4 +

8h

3

where h∈ (4,∞)

A0(h) = 8(h

2_−8h−32)

3(h−4)2 A

00_{(h) =} 256

(h−4)3

CNs: h=4+4√3 (h=4−4√3<0)

A00(4+4√3) = 256 (4√3)3 >0

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Min A(h) =

32h h−4 +

8h

3

where h∈ (4,∞)

A0(h) = 8(h

2_−8h−32)

3(h−4)2 A

00_{(h) =} 256

(h−4)3

CNs:

h=4+4√3 (h=4−4√3<0)

A00(4+4√3) = 256 (4√3)3 >0

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Min A(h) =

32h h−4 +

8h

3

where h∈ (4,∞)

A0(h) = 8(h

2_−8h−32)

3(h−4)2 A

00_{(h) =} 256

(h−4)3

CNs: h=4+4√3

(h=4−4√3<0)

A00(4+4√3) = 256 (4√3)3 >0

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Min A(h) =

32h h−4 +

8h

3

where h∈ (4,∞)

A0(h) = 8(h

2_−8h−32)

3(h−4)2 A

00_{(h) =} 256

(h−4)3

CNs: h=4+4√3 (h=4−4√3<0)

A00(4+4√3) = 256 (4√3)3 >0

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Min A(h) =

32h h−4 +

8h

3

where h∈ (4,∞)

A0(h) = 8(h

2_−8h−32)

3(h−4)2 A

00_{(h) =} 256

(h−4)3

CNs: h=4+4√3 (h=4−4√3<0)

A00(4+4√3) = 256 (4√3)3 >0

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

Min A(h) =

32h h−4 +

8h

3

where h∈ (4,∞)

A0(h) = 8(h

2_−8h−32)

3(h−4)2 A

00_{(h) =} 256

(h−4)3

CNs: h=4+4√3 (h=4−4√3<0)

A00(4+4√3) = 256 (4√3)3 >0

A cardboard poster containing 32 sq. in. of printed region is to have a margin of 2 in. at the top and bottom and 4/3 in. at each side. Determine the dimensions of the smallest piece of cardboard that can be used to make the poster.

A(h)has a relative minimum at h=4+4√3 and it is the ONLY relative extremum in (4,∞).

A(h)has an absolute minimum ath=4+4√3.

Therefore, the dimensions of smallest piece of cardboard that can be used to make a 32 sq. in. printed region is

h=4+4√3 inches andl= 8(1+4 p

(3))

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f0(x)>0ifx∈(−∞,−2)∪(−2, 0)

f0(x)<0ifx∈(0, 2)∪(2,∞)

f0(x) =0ifx=0

f0(x)is undefined whenx=−2, 2

Thus

f is increasing ifx∈(−∞,−2)∪(−2, 0)

f is decreasing ifx∈(0, 2)∪(2,∞)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f0(x)>0ifx∈(−∞,−2)∪(−2, 0)

f0(x)<0ifx∈(0, 2)∪(2,∞)

f0(x) =0ifx=0

f0(x)is undefined whenx=−2, 2

Thus

f is increasing ifx∈(−∞,−2)∪(−2, 0)

f is decreasing ifx∈(0, 2)∪(2,∞)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f0(x)>0ifx∈(−∞,−2)∪(−2, 0)

f0(x)<0ifx∈(0, 2)∪(2,∞)

f0(x) =0ifx=0

f0(x)is undefined whenx=−2, 2

Thus

f is increasing ifx∈(−∞,−2)∪(−2, 0)

f is decreasing ifx∈(0, 2)∪(2,∞)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f0(x)>0ifx∈(−∞,−2)∪(−2, 0)

f0(x)<0ifx∈(0, 2)∪(2,∞)

f0(x) =0ifx=0

f0(x)is undefined whenx=−2, 2

Thus

f is increasing ifx∈(−∞,−2)∪(−2, 0)

f is decreasing ifx∈(0, 2)∪(2,∞)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f0(x)>0ifx∈(−∞,−2)∪(−2, 0)

f0(x)<0ifx∈(0, 2)∪(2,∞)

f0(x) =0ifx=0

f0(x)is undefined whenx=−2, 2

Thus

f is increasing ifx∈(−∞,−2)∪(−2, 0)

f is decreasing ifx∈(0, 2)∪(2,∞)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f0(x)>0ifx∈(−∞,−2)∪(−2, 0)

f0(x)<0ifx∈(0, 2)∪(2,∞)

f0(x) =0ifx=0

f0(x)is undefined whenx=−2, 2

Thus

f is increasing ifx∈(−∞,−2)∪(−2, 0)

f is decreasing ifx∈(0, 2)∪(2,∞)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f00(x)>0ifx∈(−∞,−2)∪(2,∞)

f00(x)<0ifx∈(−2, 0)∪(0, 2)

f00(x) =0ifx=0

f00(x) is undefined ifx=−2, 2

Thus

f is conc. up ifx∈(−∞,−2)∪(2,∞)

f is conc. down ifx∈(−2, 0)∪(0, 2)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f00(x)>0ifx∈(−∞,−2)∪(2,∞)

f00(x)<0ifx∈(−2, 0)∪(0, 2)

f00(x) =0ifx=0

f00(x) is undefined ifx=−2, 2

Thus

f is conc. up ifx∈(−∞,−2)∪(2,∞)

f is conc. down ifx∈(−2, 0)∪(0, 2)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f00(x)>0ifx∈(−∞,−2)∪(2,∞)

f00(x)<0ifx∈(−2, 0)∪(0, 2)

f00(x) =0ifx=0

f00(x) is undefined ifx=−2, 2

Thus

f is conc. up ifx∈(−∞,−2)∪(2,∞)

f is conc. down ifx∈(−2, 0)∪(0, 2)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f00(x)>0ifx∈(−∞,−2)∪(2,∞)

f00(x)<0ifx∈(−2, 0)∪(0, 2)

f00(x) =0ifx=0

f00(x) is undefined ifx=−2, 2

Thus

f is conc. up ifx∈(−∞,−2)∪(2,∞)

f is conc. down ifx∈(−2, 0)∪(0, 2)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f00(x)>0ifx∈(−∞,−2)∪(2,∞)

f00(x)<0ifx∈(−2, 0)∪(0, 2)

f00(x) =0ifx=0

f00(x)is undefined ifx=−2, 2

Thus

f is conc. up ifx∈(−∞,−2)∪(2,∞)

f is conc. down ifx∈(−2, 0)∪(0, 2)

Let f be a continuous function over R. Describe the graph of f such that f0 is drawn as follows:

f00(x)>0ifx∈(−∞,−2)∪(2,∞)

f00(x)<0ifx∈(−2, 0)∪(0, 2)

f00(x) =0ifx=0

f00(x)is undefined ifx=−2, 2

Thus

f is conc. up ifx∈(−∞,−2)∪(2,∞)

f is conc. down ifx∈(−2, 0)∪(0, 2)