Grab bag solutions: Week 5
Topics: definition of derivative; tangent lines; higher derivatives; concavity; derivative of exponentials; product rule; chain rule.
Note: We have learned and will learn many rules for computing derivatives of functions. There
are lots of problems (which you should do!) in the book that will give you practice with applying
those rules. The purpose of this week’s Grab Bag is not to give you practice with applying the
derivative rules. Rather, the purpose of the Grab Bag is to complement the type of questions that
you can find in the book, not to replace them. So, be sure that you can also do the questions from
the book!
√
Problem 1. Suppose f (x) = 25 − x.
1. Compute f 0 (x).
2. Find the tangent line to the graph when x = 16.
3. Write an equation for the tangent line to the graph y = f (x) at the point (a, f (a)). (So, your
answer should depend on a.)
4. Suppose the tangent line to the graph of y = f (x) at the point (a, f (a)) intersects the x-axis
at the point (60, 0). What is a?
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Solution to Problem 1. The derivative is (however you want to compute it) f 0 (x) = − 2√25−x
.
√
1
Evaluating when x = 16 we get f 0 (16) = − 2√25−16
= − 16 . When x = 16, f (x) = f (16) = 9 = 3.
Thus the tangent line in question has slope −1/6 and goes through the point (16, 3), so has equation
y = − 16 (x − 16) + 3.
Plugging in x = a to f 0 (x), we get that the tangent line has equation
√
1
y=− √
(x − a) + 25 − a.
2 25 − a
If the above tangent line intersects the x-axis at the point (60, 0), then that means
√
1
0=− √
(60 − a) + 25 − a.
2 25 − a
(Plugging in 0 for y and 60 for x.) We can solve this equation for a. We get
√
1
√
(60 − a) = 25 − a
2 25 − a
and thus
1
(60 − a) = 25 − a.
2
Solving yields a = −5.
Problem 2. Suppose f (x) and g(x) are two differentiable functions. Assume that the tangent line
to the graph y = f (x) when x = 4 has equation y = 7(x − 4) + 3 and the tangent line to the graph
of g(x) when x = 4 has equation y = 5(x − 4) − 13. What are the equations for the tangent line to
the graph of y = f (x) + g(x) when x = 4 and y = f (x)g(x) when x = 4?
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Solution to Problem 2. We can deduce from the statement of the question the following information:
• f (4) = 3;
• f 0 (4) = 7;
• g(4) = −13;
• g 0 (4) = 5;
Thus
• f (4) + g(4) = −10;
•
d
dx
(f (x) + g(x)) |x=4 = f 0 (4) + g 0 (4) = 12;
• f (4)g(4) = 3 · (−13) = −39;
•
(f (x)g(x)) |x=4 = (f 0 (x)g(x) + f (x)g 0 (x)) |x=4 = f 0 (4)g(4) + f (4)g 0 (4), which is (7)(−13) +
(3)(5) = −76.
d
dx
Therefore, the tangent line to y = f (x) + g(x) when x = 4 has equation y = 12(x − 4) − 10 and the
tangent line to y = f (x)g(x) when x = 4 has equation y = −76(x − 4) − 39.
Problem 3. Suppose g(x) is defined by
(
4x2 + 3 if x < −1
g(x) =
.
8|x| − 1 if x ≥ −1
g(−1 + h) − g(−1)
from the definition of the function g.
h
h→0+
g(−1 + h) − g(−1)
2. Compute lim
from the definition of the function g.
−
h
h→0
3. Is g differentiable at x = −1? Why or why not?
1. Compute lim
Solution to Problem 3. If h > 0 and tiny, then −1 + h is negative so | − 1 + h| = −(−1 + h). Hence
g(−1 + h) = 8| − 1 + h| − 1 = −8(−1 + h) − 1 = −8h + 7
and g(−1) = 7. Thus
lim
h→0+
g(−1 + h) − g(−1)
(−8h + 7) − 7
= lim
= −8.
+
h
h
h→0
If h < 0, then
g(−1 + h) = 4(−1 + h)2 + 3 = 4(h2 − 2h + 1) + 3 = 4h2 − 8h + 7
and g(−1) = 7. Hence
lim
h→0−
(4h2 − 8h + 7) − 7
4h2 − 8h
g(−1 + h) − g(−1)
= lim
= lim
= −8.
h
h
h
h→0−
h→0−
2
Since the one-sided limits each yield −8, i.e., because
lim
h→0+
g(−1 + h) − g(−1)
g(−1 + h) − g(−1)
= −8 = lim
,
−
h
h
h→0
the (two-sided) limit
lim
h→0
g(−1 + h) − g(−1)
= −8.
h
In particular, this two-sided limit exists, so g is in fact differentiable at x = −1 (with derivative
equal to −8).
Problem 4. Suppose f is function that is concave down, and for which f 0 (x) is positive for all x.
Which of the following scenarios could truthfully describe the values of f and f 0 ?
1. f (1) = 12, f (2) = 9, f 0 (2) = −2, f (3) = 8;
2. f (1) = 10, f (2) = 14, f 0 (2) = 5, f (3) = 17;
3. f (1) = 1, f (2) = 5, f 0 (2) = 3.5, f (3) = 8;
4. f (1) = −3, f (2) = 3, f 0 (2) = 4.5, f (3) = 8.
Solution to Problem 4. Suppose a, b, c are real numbers, and a < b < c. Since f 0 (x) positive for all
x, that means f is increasing, so f (a) < f (b) < f (c). Since f is concave down, that means that the
slopes of the tangent and secant lines are ordered in this way:
f 0 (a) ≥
f (b) − f (a)
f (c) − f (b)
≥ f 0 (b) ≥
≥ f 0 (c).
b−a
c−b
Applying the facts about the values of the function, derivatives, and slopes of secant lines to the
case a = 1, b = 2, c = 3, the only choice that works is case (3) above. Thus case (3) is the only
scenario that could truthfully describe the values of f and f 0 .
Problem 5. Suppose f (x) = e−x , g(x) = 3x − 2, and set h(x) = f (x) − g(x). On your midterm,
you computed h(0) = 3 > 0 and h(1) = 1e − 1 < 0, so by that by the intermediate value theorem,
there is some c between 0 and 1 so that h(c) = 0.
1. Compute h0 (x).
2. Is h0 (x) always positive, always negative, or sometimes positive/sometimes negative?
3. On your midterm, you saw that there is no negative number x so that h(x) = 0. Using your
answer from part 2, explain why the number c is actually the only real number (positive or
negative) for which h(c) = 0.
Solution to Problem 5. We compute
h0 (x) =
d −x
e − 3x + 2 = −e−x − 3.
dx
Since e−x is always positive, no matter what x is, h0 (x) = −e−x − 3 is always negative. Thus h is a
decreasing function. Therefore, if x < c, then h(x) > h(c) = 0, while if x > c, then h(x) < h(c) = 0.
So c is only real number satisfying h(x) = 0.
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Problem 6. Suppose g is differentiable, and is an even function. Remember that g being even
means that g(−x) = g(x) for all x.
1. If you know g 0 (3) = 8, can you determine g 0 (−3)?
2. How about g 0 (0)?
Solution to Problem 6. Thinking about the graph of an even function, you can maybe guess that
g 0 (−3) = −8 and g 0 (0) = 0. This is true. In fact, g 0 (−x) = −g 0 (x) for all x (so that g 0 is an odd
function). Now that we’ve discussed the chain rule, you know that
d
d
(g(x)) =
(g(−x)) = −g 0 (−x).
dx
dx
(You can also check that g 0 (x) = −g 0 (−x) directly from the limit definition of the derivative.) Thus
g 0 (−3) = −g 0 (−(−3)) = −g 0 (3) = −8. For g 0 (0), we try the same thing: g 0 (0) = −g 0 (−0) = −g 0 (0).
Thus 2g 0 (0) = 0, so g 0 (0) = 0.
g 0 (x) =
The following problem is more abstract than we would ask you on a midterm, but nonetheless
highlights a really important concept from Calculus.
Problem 7. If h(x) is any differentiable function, we know that the tangent line to the graph of h
at the point (c, h(c)) has equation y = h0 (c)(x − c) + h(c). Define a linear function Lh,c (x) by the
formula
Lh,c (x) := h0 (c)(x − c) + h(c).
Thus the tangent line to y = h(x) at the point (c, h(c)) has equation y = Lh,c (x).
Now suppose f and g are two differentiable functions. We know that the tangent line to the graph
of f at the point (a, f (a)) has equation
y = f 0 (a)(x − a) + f (a) = Lf,a (x)
and the tangent line to the graph of g at the point (f (a), g(f (a))) has equation
y = g 0 (f (a))(x − f (a)) + g(f (a)) = Lg,f (a) (x).
1. Using the chain rule, write down a formula for the linear function Lh,a if h(x) = g(f (x)).
2. Compute the composition of the two linear functions Lg,f (a) ◦ Lf,a .
What do you notice?
Solution to Problem 7. The chain rule tells us h0 (a) = g 0 (f (a))f 0 (a). Thus
Lh,a = h0 (a)(x − a) + h(a) = g 0 (f (a))f 0 (a)(x − a) + g(f (a)).
Now we perform the composition Lg,f (a) ◦ Lf,a . We get
Lg,f (a) ◦ Lf,a (x) = g 0 (f (a)) (Lf,a (x) − f (a)) + g(f (a))
= g 0 (f (a)) f 0 (a)(x − a) + f (a) − f (a) + g(f (a))
= g 0 (f (a))f 0 (a)(x − a) + g(f (a))
= Lh,a (x).
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Thus the chain rule says that
Lg◦f,a = Lh,a = Lg,f (a) ◦ Lf,a .
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