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Exercise 13

Avg difficulty 2.5

Let pp be a function whose second derivative is p′′(x)=(x+1)(x−2)e−xp''(x) = (x+1)(x-2)e^{-x}.

  1. Construct a second derivative sign chart for pp and determine all inflection points of pp.

  2. Suppose you also know that x=5−12x = \frac{\sqrt{5}-1}{2} is a critical number of pp. Does pp have a local minimum, local maximum, or neither at x=5−12x = \frac{\sqrt{5}-1}{2}? Why?

  3. If the point (2,12e2)(2, \frac{12}{e^2}) lies on the graph of y=p(x)y = p(x) and p′(2)=−5e2p'(2) = -\frac{5}{e^2}, find the equation of the tangent line to y=p(x)y = p(x) at the point where x=2x = 2. Does the tangent line lie above the curve, below the curve, or neither at this value? Why?

Boelkins, Active Calculus, §3.3 Using derivatives to identify extreme values, Exercise 13

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.