MATH 135 – Calculus 1 Sample Questions for Exam 2 – Answers and Solutions October 10, 2016

1. An object moves along a straight line path with position given by x(t) = 4t^{2} + t − 7, (t in seconds, x in feet).

(a) What is the average velocity of the object over the interval [0, 5] of t-values?

Solution: The average velocity is:

v_{ave} = \frac{x(5) − x(0)}{5 − 0} = \frac{98 − (−7)}{5} = 21 ft/sec

(b) Fill in the following table with average velocities computed over the indicated intervals. Using this information, estimate the instantaneous velocity at t = 2.

Solution:

interval [2, 2.5] [2, 2.05] [2, 2.005] [2, 2.0005]
ave.vel.  19.0  17.2  17.02  17.002

It looks as though the average velocity is tending to 17 as the length of the interval goes to 0.

(c) Construct a similar table for intervals ending at t = 2 and repeat the calculations in the previous part. If you estimate the instantaneous velocity at t = 2 using this new information, does your result agree with what you did before (it should!)

Solution:

interval [1.5, 2] [1.95, 2] [1.995, 2] [1.9995, 2]
ave.vel.  15.0  16.8  16.98  16.998

It looks as though the average velocity is tending to 17 as the length of the interval goes to 0 again.

2. (a) What is the slope of the secant line to the graph y = x^{3} + 1 through the points with x = 1 and x = 2?

Solution: The slope of the secant line is

m_{sec} = \frac{(2^{3} + 1) − (1^{3} + 1)}{2 − 1} = \frac{9 − 2}{1} = 7.

(b) What is the slope of the secant line to the graph y = x^{3} + 1 through the points with x = 1 and x = 1 + h for a general h?

Solution: The slope of the secant line is

m_{sec} = \frac{(1 + h)^{3} + 1 − (1^{3} + 1)}{1 + h − 1} = \frac{3h + 3h^{2} + h^{3}}{h} .

(c) The slope of the tangent line to y = x^{3} + 1 at x = 1 would be obtained from what limit? Solution: The slope of the tangent is

m_{tan} = lim_{h→0} \frac{3h + 3h^{2} + h^{3}}{h} .

(d) Estimate the limit in the previous part numerically (as in the first question). Solution:

\begin{matrix}h & 1 & .5 & .05 & .005 \\ (3h + 3h^{2} + h^{3})/h & 7 & 4.75 & 3.1525 & 3.015025 \end{matrix}

It seems the limit as h → 0 is about 3.

(e) Evaluate the limit exactly using our algebraic techniques.

Solution:

lim_{h→0} \frac{3h + 3h^{2} + h^{3}}{h} = lim_{h→0} \frac{h(3 + 3h + h^{2})}{h} = lim_{h→0} 3 + 3h + h^{2} = 3

(cancelling the h’s top and bottom, then using the continuity of 3 + 3h + h^{2} as a function of h).

3. The graph of a function f is shown below with several points marked. Find all the marked points at which the following are true, and give explanations for your answers.

A  B  C D E  F  G

(a) f has an infinite discontinuity – Answer: C (from the left)

(b) f has a jump discontinuity (and: If you had a vertical axis with actual y-values to use, how would you find the one-sided limits at each such point?) – Answer: G (The other part of the question: To get the one sided limits, you would try to match the y-coordinates of the left and right sides of the graph with the axis scale. Then the limit lim_{x→G^{−}} f(x) would be the limiting y-coordinate on the left and lim_{x→G^{+}} f(x) would be the limiting value from the right.)

(c) f has a removable discontinuity – Answer: B, because lim_{x→B} f(x) exists, but is different from f(B).

(d) f is continuous – Answer: f is continuous at points A, D, E, F (and all the nonmarked points too)

4. Compute the indicated limits. Show all work for full credit.

(a) lim_{x→1} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4}

Solution: This is not an indeterminate form, and the denominator is not zero at 1, so the answer can be found by continuity:

lim_{x→1} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4} = \frac{−4}{1} = −4.

(b) lim_{x→2} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4}

Answer: This is a 0/0 form. To try to evaluate, we aim to factor the bottom and cancel factors of x − 2:

lim_{x→2} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4} = lim_{x→2} \frac{(x − 2)(3x + 1)}{(x − 2)(x − 2)} = lim_{x→2} \frac{3x + 1}{x − 2}.

This is not indeterminate any more, but the bottom is still 0 at x = 2. So this limit does not exist. (The function has an infinite discontinuity at x = 2.)

(c) lim_{x→∞} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4}

Solution: For this one, we multiply the top and bottom by \frac{1}{x^{2}} :

lim_{x→∞} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4} = lim_{x→∞} \frac{(3x^{2} − 5x − 2)\frac{1}{x^{2}}}{(x^{2} − 4x + 4)\frac{1}{x^{2}}}
= lim_{x→∞} \frac{3 − \frac{5}{x} − \frac{2}{x^{2}}}{1 − \frac{4}{x} + \frac{4}{x^{2}}}
= \frac{3 − 0 − 0}{1 − 0 + 0} = 3.

(d) lim_{x→2} \frac{\frac{1}{x^{2}} − \frac{1}{4}}{x − 2}

Solution:

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

(e) lim_{t→0} \frac{sin(6t)}{sin(7t)}

Solution: For this one, we need to use the formula lim_{u→0} \frac{sin(u)}{u} = 1 from Section 2.6. We have

lim_{t→0} \frac{sin(6t)}{sin(7t)} = lim_{t→0} \frac{6 · sin(6t)/(6t)}{7 · sin(7t)/(7t)}
= 6/7.

(f) lim_{h→0} \frac{\sqrt{h + 9} − \sqrt{9}}{h}

Solution: Multiply top and bottom by the conjugate radical:

lim_{h→0} \frac{\sqrt{h + 9} − \sqrt{9}}{h} = lim_{h→0} \frac{(\sqrt{h + 9} − \sqrt{9})(\sqrt{h + 9} + \sqrt{9})}{h(\sqrt{h + 9} + \sqrt{9})}
= lim_{h→0} \frac{(h + 9) − 9}{h(\sqrt{h + 9} + \sqrt{9})}
= lim_{h→0} \frac{h}{h(\sqrt{h + 9} + \sqrt{9})}
= lim_{h→0} \frac{1}{(\sqrt{h + 9} + \sqrt{9})}
= \frac{1}{2\sqrt{9}} = \frac{1}{6}.

5. Suppose you know each of the following conditions. What can you say about lim_{x→c} f(x) for the indicated c? Why?

(a) x^{2} + x ≤ f(x) ≤ x^{3} + 3 for all real x, at c = 0.

Solution: f(x) is not “squeezed” in this case because l(x) = x^{2} + x has the value l(0) = 0, which is strictly less than u(0) = 3 for u(x) = x^{3} + 3. The limit of f(x) as x → 0 could be any number between 0 and 3, or it might not exist at all.

(b) −x^{2} + 2x ≤ f(x) ≤ x^{4} − 4x^{3} + 6x^{2} − 4x + 2 for all real x, at c = 1 Solution: Now we have l(1) = 1 and u(1) = 1. So f(x) is “squeezed” and the Squeeze Theorem says lim_{x→1} f(x) = 1 also.

(c) f(x) = x sin (\frac{1}{x}) for all real x ≠ 0, at c = 0.

Solution: This one is trickier because we are not given l(x) and u(x). However, note that for all x ≠ 0, −1 ≤ sin (\frac{1}{x}) ≤ 1. Therefore, −x ≤ f(x) ≤ x for all x ≠ 0. Hence the Squeeze Theorem does apply and we can see lim_{x→0} f(x) = 0. (This f(x) has a graph that looks like the picture from Exercise 56 B on page 71.)

(a) What is the definition of the derivative of a function f at x = a in its domain?

Solution: f′(a) = lim_{h→0} \frac{f(a + h) − f(a)}{h}, if the limit exists. The limit lim_{x→a} \frac{f(x) − f(a)}{x − a}

also computes f′(a). These can be seen to be equivalent if you let x − a = h in the second form.

(a) Using the definition (not the shortcut rules), find f′(x) for f(x) = 3x^{3} − 2x^{2} + 1 at a general x = a.

(b) Solution:

f′(a) = lim_{h→0} \frac{(3(a + h)^{3} − 2(a + h)^{2} + 1 − (3a^{3} − 2a^{2} + 1)}{h}
= lim_{h→0} \frac{3a^{3} + 9a^{2}h + 9ah^{2} + 3h^{3} − 2a^{2} − 4ah − 2h^{2} + 1 − 3a^{3} + 2a^{2} − 1}{h}
= lim_{h→0} \frac{(9a^{2} − 4a + 9ah − 2h + 3h^{2})h}{h}
= lim_{h→0} 9a^{2} − 4a + 9ah + 2h + 3h^{2}
= 9a^{2} − 4a.

(c) Using the definition (not the shortcut rules), find f′(x) for f(x) = x^{1/2} at a general x = a > 0. Solution:

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

(b) Using the definition (not the shortcut rules), find f′(x) for f(x) = \frac{1}{x^{2}} at a general

x = a ≠ 0.

f′(a) = lim_{h→0} \frac{\frac{1}{(a+h)^{2}} − \frac{1}{a^{2}}}{h}
= lim_{h→0} \frac{a^{2} − (a + h)^{2}}{ha^{2}(a + h)^{2}}
= lim_{h→0} \frac{−2ah − h^{2}}{ha^{2}(a + h)^{2}}
= lim_{h→0} \frac{−2a − h}{a^{2}(a + h)^{2}}
= \frac{−2a}{a^{4}} = \frac{−2}{a^{3}}.