Q. 3.84 - Calculus

Question

In Exercises 83–86, use the given derivative ${\mathrm{f}}^{\text{'}}$to find any local extrema and inflection points of f and sketch a possible graph without first finding an formula for $\mathrm{f}$ .

$\mathrm{f}\text{'}\left(\mathrm{x}\right)={\mathrm{x}}^4-1$

Step-by-Step Solution

Answer

A local minimum point is $\mathrm{x}=1$ while a local maximum point is $\mathrm{x}=-1$.

Step 1: Given Information

The function's name is $\mathrm{f}\text{'}\left(\mathrm{x}\right)={\mathrm{x}}^4-1$

Step 2: Calculation

Let us consider, the following considerations when looking for the important points

 $$\begin{gathered} \mathrm{f}\text{'}\left(\mathrm{x}\right)=0 \\ {\mathrm{x}}^4-1=0 \\ \begin{array}{l}\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\left({\mathrm{x}}^2+1\right)=0\\ \Rightarrow \mathrm{x}=1,-1\end{array} \end{gathered}$$

As a result, the crucial number is $\mathrm{x}=1,-1$

Then take a look at the table below.

Interval	$\left(-\infty ,-1\right)$	$\left(-1,1\right)$	$\left(1,\infty \right)$
k	-2	0	2
Test value for ${\mathrm{f}}^{\text{'}}\left(\mathrm{k}\right)$	+	-	+
Sign of ${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)$	${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)>0$	${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)<0$	${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)>0$
Conclusion	increasing	decreasing	increasing

Therefore, A local minimum point is $\mathrm{x}=1$ while a local maximum point is $\mathrm{x}=-1.$

Then there's the graph of

L