2x2 Using the second derivative test for critical points of the function y = –x' 3 16x, which of the following statements is + true? A At x = 2, y" is positive, concave downwards, hence y(2) is a relative maximum value. At x = -4, y" is negative, concave downwards, hence y(-4) is a relative maximum value. At x = 2, y" is positive, concave upwards, hence y(2) is a relative maximum value. At x = -4, y" is negative, concave downwards, hence y(-4) is a relative minimum value.
2x2 Using the second derivative test for critical points of the function y = –x' 3 16x, which of the following statements is + true? A At x = 2, y" is positive, concave downwards, hence y(2) is a relative maximum value. At x = -4, y" is negative, concave downwards, hence y(-4) is a relative maximum value. At x = 2, y" is positive, concave upwards, hence y(2) is a relative maximum value. At x = -4, y" is negative, concave downwards, hence y(-4) is a relative minimum value.
Chapter3: Functions
Section3.3: Rates Of Change And Behavior Of Graphs
Problem 3SE: How are the absolute maximum and minimum similar to and different from the local extrema?
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Transcribed Image Text:Using the second derivative test for critical points of the function y = -x' + 2x² - 16x, which of the following statements is
3
true?
A) At x = 2, y" is positive, concave downwards, hence y(2) is a relative maximum value.
В
At x = -4, y" is negative, concave downwards, hence y(-4) is a relative maximum value.
c) At x = 2, y" is positive, concave upwards, hence y(2) is a relative maximum value.
At x = -4, y" is negative, concave downwards, hence y(-4) is a relative minimum value.
Ouectien
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