Solve the problem.
-Suppose the derivative of the function $y = f ( x )$ is $y ^ { \prime } = ( x - 3 ) ^ { 2 } ( x + 5 )$. At what points, if any, does the graph of $f$ have a local minimum or local maximum?
A) local maximum at x = -5
B) no local minimum or local maximum
C) local minimum at x = -5
D) local minimum at x = 3

Question

Quizplus · Accepted Answer

To determine the local minimum or maximum, we first find the critical points by setting the derivative equal to zero: $y' = (x - 3)^2(x + 5) = 0$. This gives us $x = 3$ and $x = -5$. To classify these critical points, we can use the first or second derivative test. Since $y' = (x - 3)^2(x + 5)$, we see that the derivative changes sign around $x = -5$ but not around $x = 3$, because the factor $(x - 3)^2$ is always non-negative. Thus, $x = -5$ is where the function has a local minimum (since the derivative changes from negative to positive), and $x = 3$ does not result in a local extremum because the derivative does not change sign (it remains non-negative due to the square).

Solve the Problem y=f(x)y = f ( x )y=f(x) Is y′=(x−3)2(x+5)y ^ { \prime } = ( x - 3 ) ^ { 2 } ( x + 5 )y′=(x−3)2(x+5)

Solve the Problem $y = f ( x )$ Is $y ^ { \prime } = ( x - 3 ) ^ { 2 } ( x + 5 )$