Question

Question 5 6 pts (05.04 MC) Which of the following functions cannot use the Second Derivative Test to determine if $x = 4$ is the location of a relative minimum or relative maximum? $y = \frac{1}{3}x^3 - 4x^2 + 16x$ $y = xe^{\frac{x}{4}}$ $y = -\cos(x - 4)$ $y = -x^2 + 8x$

          Question 5
6 pts
(05.04 MC)
Which of the following functions cannot use the Second Derivative Test to determine if $x = 4$ is the
location of a relative minimum or relative maximum?
$y = \frac{1}{3}x^3 - 4x^2 + 16x$
$y = xe^{\frac{x}{4}}$
$y = -\cos(x - 4)$
$y = -x^2 + 8x$

Question 5
6 pts
(05.04 MC)
Which of the following functions cannot use the Second Derivative Test to determine if x = 4 is the
location of a relative minimum or relative maximum?
y = (1)/(3)x^3 - 4x^2 + 16x
y = xe^(x)/(4)
y = -cos(x - 4)
y = -x^2 + 8x

Added by Jaime L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ishana K

01/28/2022

Step 1

The test states that if the second derivative is positive at a critical point, then the function has a relative minimum at that point. If the second derivative is negative at a critical point, then the function has a relative maximum at that point. Show more…

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Question 5 6 pts (05.04MC) Which of the following functions cannot use the Second Derivative Test to determine if x = 4 is the location of a relative minimum or relative maximum? 13-42+16x 3 A y=xe 4 O y=-cos(x -4) O y= -x2 + 8x