kirstenf9335jv

2023-02-18

How to use the second derivative test to find all relative extrema of $f\left(x\right)=5+3{x}^{2}-{x}^{3}$?

Vincent Burke

Beginner2023-02-19Added 9 answers

The first derivative is $f\prime \left(x\right)=6x-3{x}^{2}=3x(2-x)$, which has roots at $x=0$ and $x=2$. These are the turning point as well as the potential sites of local extrema.

Since the second derivative is $f\prime \prime \left(x\right)=6-6x$, we get $f\prime \prime \left(0\right)=6>0$ and $f\prime \prime \left(2\right)=-6<0$. The fact that $f\prime \prime \left(0\right)>0$ (and the fact that $f\prime \prime$ is continuous) implies that the graph of $f$ is concave up near $x=0$, making, by the Second Derivative Test, $x=0$ the location of a local minimum.

The fact that $f\prime \prime \left(2\right)<0$ (and the fact that $f\prime \prime$ is continuous) implies that the graph of $f$ is concave down near $x=2$, making, by the Second Derivative Test, $x=2$ the location of a local maximum.

The local minimum value (output) is $f\left(0\right)=5$ and the local maximum value (output) is $f\left(2\right)=5+12-8=9$.

Since the second derivative is $f\prime \prime \left(x\right)=6-6x$, we get $f\prime \prime \left(0\right)=6>0$ and $f\prime \prime \left(2\right)=-6<0$. The fact that $f\prime \prime \left(0\right)>0$ (and the fact that $f\prime \prime$ is continuous) implies that the graph of $f$ is concave up near $x=0$, making, by the Second Derivative Test, $x=0$ the location of a local minimum.

The fact that $f\prime \prime \left(2\right)<0$ (and the fact that $f\prime \prime$ is continuous) implies that the graph of $f$ is concave down near $x=2$, making, by the Second Derivative Test, $x=2$ the location of a local maximum.

The local minimum value (output) is $f\left(0\right)=5$ and the local maximum value (output) is $f\left(2\right)=5+12-8=9$.