Q. 29 - Calculus

Question

Use the second-derivative test to determine the local extrema of each function $f$in Exercises $29-40$. If the second-derivative test fails, you may use the first-derivative test. Then verify your algebraic answers with graphs from a calculator or graphing utility. (Note: These are the same functions that you examined with the first-derivative test in Exercises $39-50$ of Section $3.2$.)

$f\left(x\right)=(x-2{)}^2(1+x)$

Step-by-Step Solution

Answer

The local maximum is at $x=0$ and local minimum is at $x=2.$

Step 1. Given information.

The given function is $f\left(x\right)=(x-2{)}^2(1+x)$.

Step 2. Critical points.

On calculating the first derivative,

$$\begin{gathered} f\left(x\right)=(x-2{)}^2(1+x) \\ =(x^2+4-4x\left)\right(1+x) \\ =\left(x^2+4-4x\right)+(x^3+4x-4x^2) \\ =x^3-3x^2+4 \\ f\text{'}(x)=3x^2-6x \\ =3x(x-2) \end{gathered}$$

The derivative is zero at points $x=0,2.$

Therefore, these are the critical points.

Step 3. Second Derivative Test.

The second derivative of the given function is,

$f\text{'}\text{'}\left(x\right)=6x-6$

Now,

$$\begin{gathered} f\text{'}\text{'}\left(0\right)=-6<0 \\ f\text{'}\text{'}\left(2\right)=12-6=6>0 \end{gathered}$$

By the second-derivative test, since $f$ is concave down at the critical point $x=0$, $f$ has a local maximum at $x=0$. Similarly, since $f$ is concave up at the critical point $x=2$ , we know that $f$ has a local minimum at $x=2.$

Step 4. Verification.

It is clear from the graph that the local extrema's are at $x=0 and x=2.$