Answers to Quiz 1 in MATH 145: Derivatives and Concavity - Prof. Jennifer Taggart, Quizzes of Mathematics

The answers to quiz 1 in math 145, a college-level mathematics course focusing on derivatives and concavity. It includes the derivatives and second derivatives of various functions, as well as the determination of increasing functions, concavity, and inflection points.

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Pre 2010

Uploaded on 03/18/2009

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MATH 145
SAMPLE QUIZ 1 ANSWERS
1. Many answers are possible. Here’s one:
2. f0(x) = ex
(1 + ex)2. For every value of x,f0(x) is positive. So, f(x) is always
increasing.
3. f0(x) = 1 + ln xand f00(x) = 1
x. For every value of x > 0, f00(x) is positive. So, f(x)
is concave up on the interval (0,).
4. f0(x) = 1
5xex2/10 and f00(x) = 1
5ex2/10(1
5x21). The second derivative is equal
to 0 at x=±5. So, these are possible inflection points. To show that these are
inflection points, set up a sign chart for f00(x) and show that concavity changes from
up to down at 5 and from down to up at 5.
5. Note that, since Nis a density, we will only consider non-negative values of N.
f0(N) = 3(100 N2)
(100 + N2)2. This derivative is equal to 0 at N= 10. Set up a sign chart
for f0(N) to show that f(N) is increasing for densities between 0 and 10.

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MATH 145

SAMPLE QUIZ 1 — ANSWERS

  1. Many answers are possible. Here’s one:
  2. f ′(x) =

e−x (1 + e−x)^2.^ For every value of^ x,^ f^

′(x) is positive. So, f (x) is always increasing.

  1. f ′(x) = 1 + ln x and f ′′(x) = (^1) x. For every value of x > 0, f ′′(x) is positive. So, f (x) is concave up on the interval (0, ∞).
  2. f ′(x) = − 1 5

xe−x^2 /^10 and f ′′(x) =^1 5

e−x^2 /^10 (^1 5

x^2 − 1). The second derivative is equal to 0 at x = ±

  1. So, these are possible inflection points. To show that these are inflection points, set up a sign chart for f ′′(x) and show that concavity changes from up to down at −

5 and from down to up at

  1. Note that, since N is a density, we will only consider non-negative values of N.

f ′(N ) =

3(100 − N 2 )

(100 + N 2 )^2. This derivative is equal to 0 at^ N^ = 10. Set up a sign chart for f ′(N ) to show that f (N ) is increasing for densities between 0 and 10.