Practice Problems 12 for 1110
December 10, 2016
Exercise 1. Find the maximum and minimum values of f on the given interval and the points where they occur.
(a) f (x) = 9 + 3x − x2 on [0, 4]
(b) f (x) = (x2 − 1)3 on [−2, 2]
Exercise 2. Find where f has local maxima and local minima.
(a) f (x) =
x
1+x2
(b) f (x) = x3 − x2 − x − 1
(c) f (x) = x +
1
x
(d) f (x) = (x2 − 3)ex
Exercise 3. For each f (x) find the following. The implied domain of f , the intervals of increase and decrease, the local minima
and maxima, the intervals of concavity.
(a) f (x) = x3 − x2 − x − 1
(b) f (x) = x4 − 4x3 + 10
(c) f (x) =
x2 +4
2x
2
(d) f (x) = e x
(e) f (x) = (x2 − 3)ex
Exercise 4. A particle is moving along a horizontal line (with positive direction being to the right) with position function given
by s(t) = 2t3 − 14t2 + 22t − 5 for time t ≥ 0. Identify the intervals of time when the particle is moving to the left and when it
is moving to the right. Identify the intervals where the velocity is increasing and when the velocity is decreasing. Interpret this
information in terms of the motion of the particle, i.e. summarize the motion in words.
Solution. These are fairly routine applications of the derivative tests. Look at example 2 and example 3 on pages 241-242, and
all of the example starting on page 245 of section 4.4.
Exercise 5. A ball is thrown directly down from a tall building with an initial velocity of 3 m/sec. Recall that all objects in
free fall accelerate at 9.8 m/sec2 . Find the velocity function v(t) and position function s(t) of the ball. (Hint: Recall corollary 2
of the Mean Value Theorem, which says that if two functions have the same derivative, then they differ by a constant.)
Solution. Since acceleration, a(t) = −9.8, is the derivative of velocity v(t), we know that v(t) = −9.8 + c for some constant
c. But we also know the initial value v(0) = −3 because the ball is thrown downwards with that initial velocity. Then
−3 = v(0) = −9.8(0) + c implies that c = −3, so that v(t) = −9.8t − 3. Since velocity is the derivative of position, s(t), we know
that s(t) = −4.9t2 − 3t + d for some other constant d. The question forgot to specify the height of the building, so let’s just say
it is 100 meters. Then 100 = s(0) = −4.9(0)2 − 3(0) + d, so d = 100 and s(t) = −4.9t2 − 3t + 100. Of course there is nothing
special about the height of 100 meters, so that if our building has height h, then s(t) = −4.9t2 − 3t + h.