1. [5 points each] Circle the number of the graph showing each of the following functions. • (a) f(x) = 3 − e^{−x} Answer: V • (b) f(x) = x^{3} − 2x^{2} − x + 2 Answer: IV • (c) f(x) = 4 sin(πx) Answer: I • (d) f(x) = 4 − 4 cos(πx) Answer: III • Note: Graph II is y = e^{−x}. V is what you get from II, if that graph is reflected across

x-axis and shifted up by 3.

2. [20 points] One of the functions given in the following table is linear and the other is exponential. Find formulas of the appropriate type for each.

\begin{matrix}x & 1 & 2 & 3 & 4 & 5 \\ f(x) & 1.2 & 0.6 & 0.3 & 0.15 & 0.075 \\ g(x) & −2.3 & −0.6 & 1.1 & 2.8 & 4.5 \end{matrix}

Solution: g(x) is linear, since the slopes of the lines between all pairs of points in the table is 1.7. For instance using the first two points: m = \frac{−0.6+2.3}{2−1} = 1.7. The equation for g is obtained by the point-slope form: y − (−2.3) = 1.7(x − 1), so y = 1.7x − 4.

Since the problem said one is exponential, that means that f(x) is the exponential one. We can find the equation for f(x) = ca^{x} as usual. From the table with x = 1, 2, ca^{1} = 1.2 and ca^{2} = 0.6 so \frac{ca^{2}}{ca} = a = \frac{0.6}{1.2} = \frac{1}{2}. Then from the first data point c\frac{1}{2} = 1.2, so c = 2.4. g(x) = \frac{2.4}{2^{x}}.

3.

• (a) [15 points] The depth of water in a tank oscillates sinusoidally once every 4 hours. The smallest depth is 2 feet and the maximum depth is 5 feet, which occurs at t = 0. Find a formula for the depth d(t) if t is the time in hours.

Solution: The amplitude is \frac{5−2}{2} = \frac{3}{2}. The vertical shift is \frac{5+2}{2} = \frac{7}{2}. Putting t = 0 at the start of the period where there is a maximum means we want to use cos. Finally the period is 4, so we get

d(t) = \frac{3}{2} cos (\frac{πt}{2}) + \frac{7}{2}

• (b) [5 points] How fast is the depth changing at t = 1.3 hours? Is it increasing or decreasing?

Solution: The question is asking for the derivative of d(t) at t = 1.3: d′(t) = −\frac{3π}{4} sin (\frac{πt}{2}), so d′(1.3) = −\frac{3π}{4} sin (\frac{1.3π}{2}) ≐ −2.1 feet per hour. Since this is negative, the depth is decreasing at t = 1.3.

MATH 131-01, Final Exam Solutions 3 4. Compute the following limits [5 points each]. Any legal method is OK.

• (a) Solution:

lim_{x→1} \frac{x^{2} + 1}{x − 3} = \frac{lim_{x→1} x^{2} + 1}{lim_{x→1} x − 3}
= \frac{2}{−2}
= −1

• (b) Solution: This limit is indeterminate of the form 0/0, so we use L’Hopital’s Rule –

twice!

lim_{x→2} \frac{x^{2} − 4x + 4}{cos(x − 2) − 1} = lim_{x→2} \frac{2x − 4}{− sin(x − 2)}
= lim_{x→2} \frac{2}{− cos(x − 2)}
= −2

• (c) This one can also be done by L’Hopital, or by some algebraic “trickery”:

lim_{x→∞} \frac{5x^{2} − x + 21}{3x^{2} + x + 1} = lim_{x→∞} \frac{5x^{2} − x + 21}{3x^{2} + x + 1} · \frac{1/x^{2}}{1/x^{2}}
= lim_{x→∞} \frac{5 − 1/x + 21/x^{2}}{3 + 1/x + 1/x^{2}}
= \frac{5}{3}

MATH 131-01, Final Exam Solutions 4 5.

• (a) [5 points] State the limit definition of the derivative. Solution:

f′(x) = lim_{h→0} \frac{f(x + h) − f(x)}{h}

(provided that this limit exists).

• (b) [10 points] Use the definition to compute f′(x) for f(x) = \sqrt{x}.

Solution:

f′(x) = lim_{h→0} \frac{\sqrt{x + h} − \sqrt{x}}{h}
= lim_{h→0} \frac{\sqrt{x + h} − \sqrt{x}}{h} · \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}
= lim_{h→0} \frac{(x + h) − x}{h(\sqrt{x + h} + \sqrt{x})}
= lim_{h→0} \frac{1}{\sqrt{x + h} + \sqrt{x}}
= \frac{1}{2\sqrt{x}}

• (c) [10 points] Find the equation of the tangent line to the graph y = \sqrt{x} at the point [4, 2]. Solution: From part B, the derivative of \sqrt{x} at x = 4 is \frac{1}{2\sqrt{4}} = \frac{1}{4}. This is the slope of the tangent line, which has the equation y − 2 = \frac{1}{4}(x − 4), or y = \frac{1}{4}x + 1.

MATH 131-01, Final Exam Solutions 5 6. Compute the following derivatives using the derivative rules. You need not simplify. [5 points each]

• (a) f(t) = t^{3} − \frac{1}{\sqrt[4]{t}} + π^{t}.

Solution: By the power and exponential rules,

f′(t) = 3t^{2} + \frac{1}{4}x^{−5/4} + π^{t} ln(π)

• (b) g(x) = \frac{x^{2} − 2}{cos(x) + 1}

Solution: By the quotient rule,

g′(x) = \frac{(cos(x) + 1)(2x) − (x^{2} − 2)(− sin(x))}{(cos(x) + 1)^{2}}

• (c) h(z) = ln(4z^{2} + 2e^{arctan(z)})

Solution: By the chain rule,

h′(z) = \frac{8z + \frac{2e^{arctan(z)}}{1+z^{2}}}{4z^{2} + 2e^{arctan(z)}}

• (d) Find \frac{dy}{dx} if x^{2}y − 2y^{3} = 3.

Solution: Differentiating implicitly, x^{2} \frac{dy}{dx} + 2xy − 6y^{2} \frac{dy}{dx} = 0. Solving for \frac{dy}{dx}, we have

\frac{dy}{dx} = \frac{−2xy}{−6y^{2} + x^{2}}

7. Consider the family of curves defined by y = f(x) = x^{4} + 2ax^{2}, where a is any fixed real number.

• (a) [10 points] Find the critical points of f, construct a sign diagram for f′(x) in the case that a < 0. Which of your critical points are local maxima and which are local minima?

Solution: We have f′(x) = 4x^{3} + 4ax = 4x(x^{2} + a). When a < 0, this has three solutions, so three critical points: x = −\sqrt{−a}, 0, \sqrt{−a}. (Note: a < 0 means −a > 0, so \sqrt{−a} does exist in the real numbers in this case.) The sign diagram should show f′ < 0 for x < −\sqrt{−a}, f′ > 0 for −\sqrt{−a} < x < 0, f′ < 0 for 0 < x < \sqrt{−a} and f′ > 0 for x > \sqrt{−a}. This means that x = ±\sqrt{−a} are local minima and x = 0 is a local maximum (First Derivative Test).

• (b) [10 points] Repeat part a, but assume now that a > 0.

Solution: As before, f′(x) = 4x^{3} + 4ax = 4x(x^{2} + a). But when a > 0, the equation x^{2} + a = 0 has no real solutions. So there is only one critial point: x = 0. f′ < 0 for x < 0 and f′ > 0 for x > 0. So x = 0 is a local minimum.

• (c) [10 points] How many different inflection points does the graph y = f(x) have if a < 0? Explain.

Solution: When a < 0, f′′(x) = 12x^{2} + 4a = 0 if x = ±\sqrt{−a/3}. The sign of f′′ changes at each, so there are two inflection points in this case.

8. [15 points] A cubical block of dry ice (solid CO_{2}) is evaporating and losing volume at the rate of 10 cm^{3}/min. How fast are the sides of cube shrinking when the block has volume 125 cm^{3}? Give the units of your answer.

Solution: Let x denote the side of the cube, so the volume is V = x^{3}. Differentiating with respect to t, we get \frac{dV}{dt} = 3x^{2} \frac{dx}{dt}. We know \frac{dV}{dt} = −10 when V = 125. At that time, x = (125)^{1/3} = 5, so

\frac{dx}{dt} = \frac{−10}{3 · 5^{2}} = \frac{−2}{15} ≐ −.13

(units: cm/min).

9. [20 points] Ship A travels along the path given by the parametric curve x = t, y = t^{2}. At the same time, ship B travels along the curve x = t, y = 4t − 5. At what time are the two ships closest to one another?

Solution: At each time t, we can find the distance between the ships using the distance formula for points in the plane:

d(t) = \sqrt{(x_{A}(t) − y_{A}(t))^{2} + (y_{A}(t) − y_{B}(t))^{2}} = \sqrt{(t − t)^{2} + (t^{2} − 4t + 5)^{2}} = |t^{2} − 4t + 5|

But note that t^{2} − 4t + 5 = (t − 2)^{2} + 1 > 0 for all t. So, we get d(t) = t^{2} − 4t + 5. We want the time when the ships are closest, so we are looking for the minimum of d(t). To find this, take d′(t) = 2t − 4 and set = 0. This says t = 2.

Note: Many people made tables of the distance at whole number values for t: t = 0, 1, 2, 3, etc., and concluded that the minimum distance ocurred at t = 2 from that information. While the conclusion is correct, the reasoning is not. How do you know that the distance doesn’t reach an even smaller value for some t between 1 and 2, or t between 2 and 3? So this solution did not receive full credit, even though the answer was correct.

10.

• (a) [7.5 points] Compute the left- and right-hand sums for the function f(t) = e^{−t^{2}} on the interval [0, 1], using n = 4 equal subdivisions.

Solution: Δt = \frac{1−0}{4} = .25.

LHS = e^{0}Δt + e^{−(.25)^{2}}Δt + e^{−(.5)^{2}}Δt + e^{−(.75)^{2}}Δt ≐ .82

and

LHS = e^{−(.25)^{2}}Δt + e^{−(.5)^{2}}Δt + e^{−(.75)^{2}}Δt + e^{−1}Δt ≐ .66

Note: A lot of people were computing values of this function incorrectly. I couldn’t tell exactly what went wrong, but I think it was probably a calculator issue. So I didn’t assess a large penalty if your method was OK, but the values were incorrect.

• (b) [7.5 points] Consider the graph y = f(t) given below. Compute the average value \overline{y} of f on the interval 0 ≤ t ≤ 5, given that the area marked A is 25 square units, and the areas marked B and C are both 16 square units.

MATH 131-01, Final Exam Solutions 8 Solution: Since the areas A and C are below the t-axis, in computing ∫_{0}^{5} f(t) dt, we count them with negative signs. So the average value is

\overline{y} = \frac{1}{5 − 0} ∫_{0}^{5} f(t) dt = \frac{1}{5}(−25 + 16 − 16) = −5.