MA 151, Applied Calculus, Spring 2020, Midterm 2 preview solutions Let f(x) = x + x and g(x) = e x + x.

1. Let f (x) =√

x + xand g(x) = e^x+ x.

(a) Find f (x)g(x) (b) Find f (x)/g(x).

(c) Find f (g(x)).

(d) Find g(f (x)).

Solution:

(a) f (x)g(x) = (√

x + x)(e^x+ x) (b) f (x)/g(x) =

√x + x e^x+ x. (c) f (g(x)) = f (e^x+ x) =√

e^x+ x + (e^x+ x).

(d) g(f (x)) = g(√

x + x) = e⁽

√x+x)+ (√ x + x).

2. Let f (x) = e^x+ e^−x. The graph of f (x) is shown to the right.

(a) Give a formula for, and identify which graph below is 2f (x). Solution: 2f (x) = 2(e^x+ e^−x), graph (v).

(b) Give a formula for, and identify which graph below is f (2x). Solution: f (2x) = e^2x+ e^−2x, graph (i).

(c) Give a formula for, and identify which graph below is f (x) + 2. Solution: f (x) + 2 = e^x + e^−x+ 2, graph (iii).

(d) Give a formula for, and identify which graph below is f (x + 2). Solution: f (x + 2) = e^x+2+ e^−(x+2), graph (ii).

(e) Give a formula for, and identify which graph below is f (x − 2). Solution: f (x − 2) = e^x−2+ e^−(x−2), graph (iv).

(i)

(ii)

(iii)

(iv) (v)

3. (a) Rewrite as a power of x: 1

√x Solution:

√1

x = x^−1/2

(b) Rewrite −x^2/3 as a radical√

or a fraction 1

, or both.

Solution:

−x^2/3= −³

√

x² = −(√³ x)²

(c) Write an equation that is equivalent to the description: w is inversely proportional to the square root of x

Solution:

w = C 1

√x

4. Let f (x) = x3^x. Find an approximation of f⁰(0.5)by filling in the table below. Fill in 6 distinct values of x, given by x = 0.5 ± 1, x = 0.5 ± 0.01 and x = 0.5 ± 0.001

Solution:

x f (x) = x3^x f (0.5) − f (x) 0.5 − x

0.400 0.620738 2.452872

0.490 0.839432 2.659345

0.499 0.863344 2.681053

0.500 0.866025 ERR

0.600 1.159909 2.938838

0.510 0.893104 2.707856

0.501 0.868711 2.685904

The two closest values for f⁰(x)are when x = 0.499 and 0.501. Averaging these together we have our best estimage

f⁰(0.5) ≈ 2.683

5. Based on the graph below, make as accurate as possible of an estimate for f⁰(−4)and f⁰(3):

Solution:

Here are the tangent lines (yes, these are unnaturally perfect)

It looks to me like the line on the left goes through (−6, 7.25) and (0, −3.5). Thus, m = f⁰(−4) ≈ 7.25 − (−3.5)

−6 − 0 = −1.79.

It looks to me like the lin on the right goes through (−2, −3) and (6, 5.5). Thus, m = f⁰(3) ≈ −3 − 6

−2 − 5.5 = 1.2.

6. Shown below are eight functions. Four of them are derivatives of the other four. Identify which function is the derivative of which other function; justify each answer by referring to a specific x-value, indicate whether the function is flat/increasing/decreasing and whether the derivative is 0/positive/negative at this x-value. For instance, you could say “(u) is the derivative of (v), because at x = 1 we have (v) is increasing and (u) is positive,” or you could be more abbreviated and say the same thing with “u=v⁰, @ x = 1, v ↑, u=+.”

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

Solution:

Make sure you don’t reverse which is the derivative and which is the original function.

• a = c⁰, @ x = 0, c ↑ and a = +. (Note: to make sure that this is the right answer you need to check more points. E.g. @ x = −4, c ↑ and a = +. @ x = 5, c ↑ and a = +.)

• f = b⁰, @ x = 1, b ↓ and f = −. (Note: to make sure that this is the right answer you need to check more points. E.g. @ x = −4, b ↓ and f = −. @ x = 5, b ↑ and f = +.)

• h = e⁰, @ x = 1, e ↑ and h = +. (Note: to make sure that this is the right answer you need to check more points. E.g. @ x = −4, e ↓ and h = −. @ x = 4, e ↓ and h = −. )

• d = g⁰, @ x = 1, g ↓ and d = −. (Note: to make sure that this is the right answer you need to check more points. E.g. @ x = −3, g ↑ and d = +. @ x = 3, g ↓ and d = −. )

Make sure you don’t reverse which is the derivative and which is

the original function!!! This is really common mistake to make!

7. Let f (x) be defined by the graph below

For each quantity below, indicate whether it is +, − or 0

f (0.75) = − f⁰(0.75) = − f⁰⁰(0.75) = 0 f (1) = − f⁰(1) = − f⁰⁰(1) = + f (1.25) = − f⁰(1.25) = 0 f⁰⁰(1.25) = + f (2.25) = + f⁰(2.25) = + f⁰⁰(2.25) = 0 f (3) = + f⁰(3) = 0 f⁰⁰(3) = − f (4) = 0 f⁰(4) = − f⁰⁰(4) = 0 f (5) = − f⁰(5) = 0 f⁰⁰(5) = + f (6) = 0 f⁰(6) = + f⁰⁰(6) = 0 f (7.75) = + f⁰(7.75) = − f⁰⁰(7.75) = 0 f (8.75) = − f⁰(8.75) = 0 f⁰⁰(8.75) = + f (9.25) = − f⁰(9.25) = + f⁰⁰(9.25) = 0

For the first column, you simply look at the y-values on the graph, and see if they are above/below/or the x-axis.

For the second column, you look at the graph, and see if the slope is up/down/0.

For the third column you look at which way the graph is curving: + means curving up or less down, − means curving down or less up, and 0 means not curving much at all (it may be easiest to see this if it’s half way between spots that are clearly curving one way and then the other).

8. Find the following derivatives (a) d

dx7x⁷ (b) d

dx3√³ x (c) d

dx6 1 x⁵ (d) d

dx5e^x (e) d

dx 1 2ln(x) Solution:

(a) d

dx7x⁷ = 49x⁶ (b) d

dx3√³ x = d

dx3x^1/3 = x^−2/3

 or 1

x^2/3

 .

(c) d dx6 1

x⁵ = −30x⁻⁶

 or −30

x⁶

 .

(d) d

dx5e^x = 5e^x (e) d

dx 1

2ln(x) = 1 2 ·1

x

 or 1

2x

 .

9. Find the derivative of each of the following functions (a) f (x) = 17

√3

x − 1.7 ln(x) + 5e^x Solution:

f⁰(x) = −17

3 x^−4/3−1.7 x + 5e^x (b) g(x) = 17

x¹⁷ − 1.7(1.8)^x Solution:

g⁰(x) = −289x⁻¹⁸− 1.7 ln(1.8)(1.8)^x

10. Find the equation of the tangent line to f (x) = −x⁴− 32 · 1

x² at the point x = 4. Simplify the numbers in your answer.

Solution:

y = m(x − x0) + y0

x0= 4 y₀= f (4)

= −(4)⁴− 32 · 1 4²

= −256 − 32 16

= −258

f⁰(x) = −4x³− 32(−2)x⁻³

= −4x³+ 64 · 1 x³ m = f⁰(4)

= −4(4)³+ 64 · 1 4³

= −256 + 64 64

= −255

y = −255(x − 4) − 258

11. Find the following derivatives, do not simplify your answers.

(a) d

dq23e^3q+1 (b) d

dx− 5 ln(5x + 11) (c) d

dt(t²+ t)¹⁰ (d) d

dx 5 11x − 7 (e) d

dq

pq²+ 1

Solution:

(a) Do it this way with y and z

y = 23e^z z = 3q + 1 dy

dq = dy dz ·dz

dq

= 23e^z(3)

= 23e^3q+1(3)

or do it this way and say the instructions in your head at each step:

d

dq23e^3q+1=23e^3q+1·(3) deriv. of

outside don’t change

inside

deriv. of inside

(b) Do it this way with y and z

y = −5 ln(z) z = 5x + 11 dy

dx = dy dz ·dz

dx

= −51 z(5)

= − 5 5x + 1(5)

or do it this way and say the instructions in your head at each step:

d

dx − 5 ln(5x + 11) = −5 1

5x + 11 ·(5) deriv. of

outside

don’t change

inside

deriv. of inside

(c) Do it this way with y and z

y = z¹⁰ z = t²+ t dy

dt = dy dz ·dz

dt

= 10z⁹(2t + 1)

= 10(t²+ t)⁹(2t + 1)

or do it this way and say the instructions in your head at each step:

d

dt(t²+ t)¹⁰= 10(t²+ t)⁹·(2t + 1) deriv. of

outside don’t change

inside

deriv. of inside

(d) Do it this way with y and z

y = −5z⁻¹ z = 11x − 7 dy

dx = dy dz ·dz

dx

= −z⁻²(11)

= −(11x − 7)⁻²(11)

or do it this way and say the instructions in your head at each step:

d

dx5(11x − 7)⁻¹ = −5(11x − 7)⁻² ·(11) deriv. of

outside don’t

change inside

deriv. of inside

(e) Do it this way with y and z

y = z^1/2 z = q²+ 1 dy

dq = dy dz ·dz

dq

= 1

2z^−1/2(2q)

= 1

2(q²+ 1)^−1/2(2q)

or do it this way and say the instructions in your head at each step:

d

dq(q²+ 1)^1/2= 1

2(q²+ 1)^−1/2·(2q)

deriv. of

outside don’t change

inside

deriv. of inside

12. Find the following derivatives, do not simplify your answers.

(a) d dq

√q ln(q)

(b) d

dx(2x²− 5x)(3e^x+ ln(x)) (c) d

dx

5 − 4x + 9x² 2 + 10x + e^x Solution:

(a)

√q

|{z}

f

ln(q)

| {z }

g

f⁰ = 1 2√

q g⁰ = 1

q f⁰g + f g⁰ = 1

2√

q ln(q) +√ q1

q (b)

(2x²− 5x)

| {z }

f

(3e^x+ ln(x))

| {z }

g

f⁰ = 4x − 5 g⁰ = 3e^x+ 1

x

f⁰g + f g⁰ = (4x − 5)(3e^x+ ln(x)) + (2x²− 5x)(3e^x+ 1 x) (c)

f

z }| {

5 − 4x + 9x² 2 + 10x + e^x

| {z }

g

f⁰ = 18x − 4 g⁰ = e^x+ 10 f⁰g − f g⁰

g² = (18x − 4)(2 + 10x + e^x) − (5 − 4x + 9x²)(e^x+ 10) (2 + 10x + e^x)²