91Ó°ÊÓ

We have the given derivative function $f^{\mathrm{′}}\left(x\right)=x^4−1$

Here we need to find the local extrema and inflection points.

Let $f$be a function that is differentiable on an interval $I$.

• If $f^{\text{'}}$is positive in the interior of $l$, then $f$is increasing on $I$.
• If $f^{\text{'}}$is negative in the interior of $I$, then $f$is decreasing on $I$.
• If $f^{\text{'}}$is zero in the interior of $l$, then $f$is constant on $I$.

Suppose both $f$and $f^{\text{'}}$are differentiable on an interval $l$, then

If $f^{\text{'}\text{'}}$is positive on $/$, then $f$is concave up on $I$.

Again

If $f^{\text{'}\text{'}}$is negative on $I$, then $f$is concave down on $I$.

Suppose $x=c$is the location of a critical point of a function $f$, and let $(a,b)$be an open interval around $c$that is contained in the domain of $f$and does not contain any other critical point of $f$. If $f$continuous on $(a,b)$and differentiable at every point of $(a,b)$except possibly at $x=c$, then

• If $f^{\text{'}}\left(x\right)$is positive for $x∈(a,c)$and negative for $x∈(c,b)$, then $f$has a local maximum at $x=c$.
• If $f^{\text{'}}\left(x\right)$is negative for $x∈(a,c)$and positive for $x∈(c,b)$, then $f$has a local minimum at $x=c$.
• If $f^{\text{'}}\left(x\right)$is positive for both $x∈(a,c)$and $x∈(c,b)$, then $f$does not have a local extremum at $x=c$.
• If $f^{\text{'}}\left(x\right)$is negative for both $x∈(a,c)$and $x∈(c,b)$, then $f$does not have a local extremum at $x=c$.

Here,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{1}{x} \\ \frac{1}{x}=0 \\ x=0 \end{gathered}$$

This derivative $f^{\text{'}}$has no critical point. It is undefined at $x=0$.

Since it is undefined at $x=0$, by the first derivative test, $f$has no local extrema.

Inflection points of a function are the points in the domain of $f$at which its concavity changes.

Since the sign of $f^{\text{'}\text{'}}$measures the concavity of $f$, you can find inflection points by looking for the places where $f^{\text{'}\text{'}}$changes sign.