Q. 30 - 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 $3.2.$)

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

Step-by-Step Solution

Answer

The local extrema's are at the points $x=0 (Local Maximum), and x=1,-1(Local Minimum).$

Step 1. Given information.

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

Step 2. Critical points .

On calculating the first derivative, 

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

The derivative is zero at points $x=0,1,-1.$

Therefore, these are the critical points. 

Step 3. Second Derivative Test.

The second derivative of the given function is, 

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=12x^2-2 \\ f\text{'}\text{'}\left(0\right)=-2<0 \\ f\text{'}\text{'}\left(1\right)=10>0 \\ f\text{'}\text{'}(-1)=10>0 \end{gathered}$$

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 $x=1,-1$ , we know that $f$ has a local minimum at $x=1,-1.$ 

Step 4. Verification.

It is clear from the graph that the local extrema's are at$x=0,1,-1.$