Doug C. answered • 12/31/20

Math Tutor with Reputation to make difficult concepts understandable

Tina S.

asked • 12/31/20The graph of the continuous function g is shown above for −3≤x≤5 . The function g is twice differentiable except at x=1

Graph:

http://prnt.sc/wdcalq

Let f be the function with f(1)=3 and derivative given by f′(x)=(6x^2−5x)e^x.

(a) Find the x -coordinate of each critical point of f . Classify each critical point as the location of a relative minimum, a relative maximum, or neither. Justify your answers.

(b) Find all values of x at which the graph of f has a point of inflection. Give reasons for your answers.

(c) Fill in the missing entries in the table below to describe the behavior of g′ and g′′ on the interval −3≤x≤5. Indicate Positive or Negative. Give reasons for your answers.

Chart: https://prnt.sc/wdcflj

(d) Let h be the function defined by h(x)=f(x)g(x) . Is h increasing or decreasing on the interval 1< x < 5? Give a reason for your answer.

Thank you

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Doug C. answered • 12/31/20

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Math Tutor with Reputation to make difficult concepts understandable

Daniel B. answered • 12/31/20

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A retired computer professional to teach math, physics

First calculate the second derivate f"

f'(x) = (6x² - 5x)e^{x} = x(6x - 5)e^{x}

f"(x) = (12x - 5)e^{x} + (6x² - 5x)e^{x} = (6x² + 7x - 5)e^{x}

(a)

The critical points of f are where f'(x) = 0, i.e.,

x = 0 and x = 5/6.

x = 0 is relative maximum because f"(0) < 0

x = 5/6 is relative minimum f"(5/6) > 0

(b)

f has inflexion point where f" is 0.

Since e^{x} is never 0, we just need to solve the quadratic equation

6x² + 7x - 5 = 0

x = -5/3 and x = 1/2

(c)

g' is positive on intervals where g is increasing, and g' is negative where g is decreasing.

g" is positive where g is convex upward and g" is negative where g is convex downward.

(d)

h(x) is increasing where its derivative is positive.

Let's consider its derivative

h'(x) = f'(x)g(x) + f(x)g'(x)

To see whether h'(x) is positive on 1 < x < 5, let's see how f, f', g, g' behave on that interval.

f'(x) > 0 on 1 < x < 5 for the following two reasons:

e^{x} is always positive

6x² - 5x is a quadratic function with both roots < 1 and its leading coefficient, 6, is positive

f(x) > 0 on 1 < x < 5 for the following reason:

f(1) = 3, as given, and f(x) is increasing after that (since f'(x) > 0)

g(x) < 0 on 1 < x < 5 by inspection of the graph

g'(x) < 0 on 1 < x < 5 because g(x) is decreasing there

Combining the above determinations for f, f', g, g',

h'(x) < 0 on 1 < x < 5 and therefore

h(x) is decreasing on 1 < x < 5.

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