## . *Consider the funion defined implicitly near (2, 1) by x* ^{2} *y* ^{2} *+ 3xy = 10y.*

## (a) [ points] *Use implicit diﬀerentiation to find the derivative at (2, 1).*

*d* *dx*

## h *x* ^{2} *y* ^{2} *+ 3xy* i

## = *d*

*dx* *[10y]*

*2xy* ^{2} *+ 2yx* ^{2} *dy*

*dx* *+ 3y + 3x* *dy*

*dx* = 10 *dy* *dx* *2yx* ^{2} *dy*

*dx* *+ 3x* *dy*

*dx* − 10 *dy*

*dx* *= −2xy* ^{2} *− 3y* *dy*

*dx* = − *2xy* ^{2} *+ 3y* *2yx* ^{2} *+ 3x − 10* *Subﬆituting x = 2 and y = 1, we get*

*dy* *dx*

## _{(2,1)} = − 4 + 3 8 + 6 − 10 *dy*

_{(2,1)}

*dx*

## _{(2,1)} = − 7 4

_{(2,1)}

## (b) [ points] *Determine the equation of the tangent line at (2, 1).*

*The equation of the tangent line is y = −* 7

## 4 *(x − 2) + 1 .*

## . *Let f (x) be the funion defined by f (x) = e* ^{√} ^{x} *for x ≥ 0.*

^{x}

## (a) [ points] *Write an equation for the line tangent to the graph of f (x) at the* *point where x = 1.*

*f* ^{′} *(x) =* *d*

*dx* *e* ^{√} ^{x} *= e* ^{√} ^{x} *d* *dx*

^{x}

^{x}

## √ *x =* 1 2 √

*x* *e* ^{√} ^{x} *,* *So the slope of the tangent line at x = 1 is f* ^{′} (1) = *e*

^{x}

## 2 . The equation for the *tangent line is y =* *e*

## 2 *(x − 1) + e .*

## (b) [ points] *Find the point where this tangent line crosses the x-axis.*

*The line crosses the x-axis when y = 0.*

*e*

## 2 *(x − 1) + e = 0* *e*

## 2 *(x − 1) = −e* *x* − 1 = −2 *x =* −1 *The tangent line crosses the x-axis at (−1,0).*

## . *Let f be the funion defined by f (x) = x* ^{3} *− x* ^{2} *− 2x for 0 ≤ x ≤ 3.*

## (a) [ points] *Find where the graph of f crosses the x-axis. Show your work.*

*Set f (x) = 0 and solve*

*x* ^{3} *− x* ^{2} *− 2x = 0* *x(x* ^{2} *− x − 2) = 0* *x(x + 1)(x* − 2) = 0

*The solutions are x = −1, x = 0, x = 2, but we are reﬆried to [0,3], so the* *only two places where f crosses the x-axis will be x = 0 and x = 2.*

## (b) [ points] *Find the intervals on which f is increasing.*

## Find the derivative and the critical points.

*f* ^{′} *(x) = 3x* ^{2} *− 2x − 2* Now solve

*3x* ^{2} *− 2x − 2 = 0* *x =* 2 ± √

## 28

## 6 = 1 ± √ 7 3 *Only x =* 1 + √

## 7

## 3 *lies in our interval. We know that f (x) is decreasing if* *f* ^{′} *(x) < 0 which happens when 0 < x <* 1 + √

## 7

## 3 *. Also, f (x) is increasing if* *f* ^{′} *(x) > 0 which happens when* 1 + √

## 7

## 3 *< x < 3.*

## (c) [ points] *Find the absolute maximum and absolute minimum value of f .* Juﬆify your answer.

*f (x) is continuous on the closed interval [0, 3], so it attains its maximum and* *minimum. There are three points to check: x = 0, x =* 1 + √

## 7

## 3 *, and x = 3.*

## Spring MATH -

## From the Firﬆ Derivative Teﬆ we can see that the absolute minimum occurs *at x =* 1 + √

## 7

## 3 *. Evaluating f at and , we have that f (0) = 0 and f (3) = 12,* *so the absolute maximum occurs at x = 3.*

*The maximum value of f on [0, 3] is and the minimum value is −* 20 + 14 √ 7

## 27 ≈

*−2.1126.*

## . [ points] A projeile is fired toward the sea from a cliﬀ feet above the water.

## If the launch angle is 45 ^{◦} *and the muzzle velocity is k, then the height h of the* *projeile above the water when it is x feet downrange from the launch site is*

*h(x) =* −32

*k* ^{2} *x* ^{2} *+ x + 200 feet.*

## See the figure.

*Determine the maximum values of h and the diﬆance downrange where this maxi-* *mum occurs if k = 2800 feet per second.*

**x**

**h**

cliff

sea level

*h(x) =* − 32

*k* ^{2} *x* ^{2} *+ x + 200* *h* ^{′} *(x) = −* 64

*k* ^{2} *x + 1* Set equal to zero and solve

## − 64

*k* ^{2} *x + 1 = 0* *x =* *k* ^{2}

## 64

*Since h* ^{′′} *(x) < 0, this will be the absolute maximum. Thus, the downrange diﬆance* *is x =* *k* ^{2}

## 64 *= 122, 500 feet. The height at that diﬆance is h(122500) = 61, 450 feet*

## Spring MATH -

## . Find the requeﬆed quantities in each of the following.

## (a) [ points] Let

*f (x) =*

##

##

##

##

##

##

##

*2 + cos(x − 1) if 0 ≤ x ≤ 1,* *ax* − 2 *if 1 < x < 2,* *x* ^{2} *− 2x + b* *if 2 ≤ x ≤ 3.*

*where a and b are conﬆants. Find the values of a and b which make f (x) is* *continuous on [0, 3].*

## We need lim

*x* →1

^{−}

*f (x) = lim*

*x* →1

^{+}

*f (x). This gives the equation* *2 + cos(0) = a − 2*

*a = 5* Likewise, we need lim

*x* →2

^{−}

*f (x) = lim*

*x* →2

^{+}

*f (x). This gives* *2a − 2 = 2* ^{2} *− 2(2) + b*

*8 = 4 − 4 + b* *b = 8*

*These values for a and b will make f (x) continuous.*

## (b) [ points] Evaluate lim

*n* →∞

*6n* ^{2} *− 3n + 7* *(n + 3)* ^{2} .

*n* lim →∞

*6n* ^{2} *− 3n + 7*

*(n + 3)* ^{2} = lim

*n* →∞

*6n* ^{2} *− 3n + 7* *n* ^{2} *+ 6n + 9*

## = lim

*n* →∞

## 6 *n* ^{2} *n* ^{2} − 3 *n*

*n* ^{2} + 7 *n* ^{2} *n* ^{2}

*n* ^{2} + 6 *n* *n* ^{2} + 9

*n* ^{2}

## = 6 1 = 6

## . Find the following limits (a) [ points] lim

*x* →0

*sin(x)* *x cos(x)* .

## Use l’Hospital’s Rule since both the numerator and denominator go to .

*x* lim →0

*sin(x)*

*x cos(x)* = lim

*x* →0

*cos(x)* *cos(x) − x sin(x)*

## = 1

## 1 − 0 = 1

## Spring MATH -

## This can also be done as follows:

*x* lim →0

*sin(x)*

*x cos(x)* = lim

*x* →0

*sin(x)* *x*

## 1 *cos(x)*

## = 1 · 1 = 1

## (b) [ points] lim

*x* →∞

*ln(xe* ^{x} )

^{x}

*x* .

## Use l’Hospital’s Rule since both the numerator and denominator go to infin- ity.

*x* lim →∞

*ln(xe* ^{x} )

^{x}

*x* = lim

*x* →∞

## 1

*xe* ^{x} *(e* ^{x} *+ xe* ^{x} ) 1

^{x}

^{x}

^{x}

## = lim

*x→∞*

*e* ^{x} *+ xe* ^{x} *xe* ^{x}

^{x}

^{x}

^{x}

## = lim

*x* →∞

## 1

*x* + 1 = 1

## (c) [ points] lim

*x* →2

*x* ^{2} − 4 *x* ^{2} *− x + 6* .

## l’Hospital’s Rule does not apply.

*x* lim →2

*x* ^{2} − 4 *x* ^{2} *− x + 6* = 0

## 8

## = 0

## . Find the following general antiderivatives.

## (a) [ points]

## Z

*x* ^{2} *+ x + x* ^{−1} *+ x* ^{−2} *dx* Z

*x* ^{2} *+ x + x* ^{−1} *+ x* ^{−2}

*dx =* 1 3 *x* ^{3} + 1

## 2 *x* ^{2} *+ ln |x| −* 1 *x* *+ C.*

## (b) [ points]

## Z

*2 sin(x) − 2cos(x) + 2sec* ^{2} *(x)* *dx* Z

*2 sin(x) − 2cos(x) + 2sec* ^{2} *(x)*

*dx =* *−2cos(x) − 2sin(x) + 2tan(x) + C.*

## (c) [ points]

## Z * e* ^{x} *− e* ^{−x} 2

^{x}

^{−x}

## ! *dx.*

## Z * e* ^{x} *− e* ^{−x} 2

^{x}

^{−x}

## !

*dx =* 1 2

## Z

*(e* ^{x} *− e* ^{−x} *)dx =* 1

^{x}

^{−x}

## 2 *(e* ^{x} *+ e* ^{−x} *) + C*

^{x}

^{−x}

## Spring MATH -

## . Use the Fundamental Theorem of Calculus (part ) to evaluate the following deriva- tives.

## (a) [ points] *If F(x) =* Z *x*

### 1

## cos ^{2} *(u)e* ^{−u}

^{−u}

^{2}

*du then F* ^{′} *(x) equals*

## By the Fundamental Theorem of Calculus, Part I, we know that *F* ^{′} *(x) =* *d*

*dx* Z *x*

### 1

## cos ^{2} *(u)e* ^{−u}

^{−u}

^{2}

*du = cos* ^{2} *(x)e* ^{−x}

^{−x}

^{2}

*.*

## (b) [ points] *If G(x) =* Z _{3x}

_{3x}

^{2}

### 1

*cos(2u) sin(2u) du then G* ^{′} *(x) equals*

## By the Fundamental Theorem of Calculus, Part I, and the Chain Rule we have

*G* ^{′} *(x) =* *d* *dx*

## Z *3x*

^{2}

### 1

*cos(2u) sin(2u) du*

## = cos

*2(3x* ^{2} ) sin

*2(3x* ^{2} ) *d* *dx* *(3x* ^{2} )

*= 6x cos(6x* ^{2} *) sin(6x* ^{2} )

## . Find the following definite integrals.

## (a) [ points]

## Z _{3}

### 1 *4x −* 1 *x* *dx*

## Z _{3}

### 1 *4x −* 1

*x* *dx = 2x* ^{2} *− ln(x)*

### 3 1

*= (18 − ln3) − (2 − ln1)*

*= 16 − ln3 ≈ 14.9014*

## (They may use the calculator to compute this integral, but in that case the answer is either right or wrong — no partial credit.)

## Spring MATH -

## (b) [ points] *The funion h(x) is graphed below.*

## 1 2 3 4 5 6 7 8 9 10

## 1 2 3 4 5 6

## Compute Z 10

### 0

*h(x) dx.*

## This is to be done geometrically.

## Z 10 0

*h(x) dx =* Z 1

### 0

*h(x) dx +* Z 2

### 1

*h(x) dx +* Z 4

### 2

*h(x) dx +* Z 5

### 4

*h(x) dx +* Z 10

### 5

*h(x) dx*

## = (0)(1) + 1

## 2 (2 + 4)(1) + 1

## 2 (4 + 2)(2) + (2)(1) + 1

## 2 (1 + 5)(5)

## = 26

## . Find the requeﬆed areas.

## (a) [ points] Set up (but do NOT evaluate) the integral that will give the area *enclosed by the curves y = cos(x), y = 1+sin(2x) and the vertical line x =* *3π*

## 2 .

### Π

Π

2

3 Π 2

### 1

### 2 3

## Area = Z

^{3π}_{2}

### 0

*[(1 + sin(x)) − cos(x)] dx*

## (The corre answer is 2 + *3π*

## 2 *≈ 6.71239. They do not have to evaluate the* integral, but IF they do, they muﬆ do it correly.)

## Spring MATH -

## (b) [ points] *The region R is enclosed by y = x/4 and y =* √

*x in the firﬆ quad-* *rant. Find the area of R.*

### 4 8 12 16

### 1 2 4 3

## Firﬆ, find the points of interseion by setting √

*x =* ^{x} _{4} *. These are x = 0 and* *x = 16. The area is then*

^{x}

## Area _{R} = Z _{16}

_{R}

### 0

## √ *x* − *x*

## 4

*dx =* 32

## 3 *≈ 10.6667.*

## (They may use the calculator to compute this integral, but they muﬆ show the integral that they are evaluating and not juﬆ the answer.)

## BONUS [ points each] Compute the derivatives of the following funions. You do not need to simplify.

## (a) *f (x) =* *x* ^{2} + 1 *e* ^{2x} .

^{2x}

*f* ^{′} *(x) =* *2xe* ^{2x} *− 2(x* ^{2} *+ 1)e* ^{2x} *e* ^{4x}

^{2x}

^{2x}

^{4x}