Q. 84 - Calculus

Question

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

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

Step-by-Step Solution

Answer

No local extrema, critical points,

Step 1. Given data

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

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

Step 2. Local extrema

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

Step 3. Critical point

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.