Question

Use the given derivative $f\text{'}$to find any local extrema and inflection points of $f$and sketch a possible graph without first finding an formula for $f$.

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

Short answer

$x=1$is a local minimum point and $x=-1$is a local maximum point.

Step-by-step solution

Step 1. Given information 

The function is given by$f\text{'}\left(x\right)=x^4-1$.

Step 2. Calculation

To find the critical points consider $f\text{'}\left(x\right)=0$

$$\begin{gathered} x^4-1=0 \\ (x-1)(x+1)(x^2+1)=0 \\ \Rightarrow x=1,-1 \end{gathered}$$

So, the critical number is $x=1,-1$

Then consider the following table,

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

So, $x=1$is a local minimum point and $x=-1$is a local maximum point.

Then the graph of $f,$