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 an formula for $f$.

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

Step-by-Step Solution

Answer

$f$ has a local minimum at$x=1$-

The function has no local maxima.

The function has inflection point at$x=0$

Step 1: Given information

The given derivative of a function is

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

Step 2: Calculation

Theorem $3.6$states that the derivative measures where a function is increasing or decreasing, let f be a function that is differentiable on an interval $I.$

(a) If $f^{\text{'}}$is positive in the interior of$l,$ then $f$is increasing on$l.$

(b) If $f^{\text{'}}$ is negative in the interior of I, then$f$ is decreasing on $I.$

(c) If $f^{\text{'}}$is zero in the interior of$I,$ then f is constant on $I.$

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

(a) 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 $x=c.$

(b) If $f^{\text{'}}(x$$)$ is negative for $x\in (a,c)$ and positive for $x\in (c,b),$then f has a local minimum at $x=c.$

(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 $x=c.$

(d) 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 oxtromum at $x=c.$

Here

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

Now simplifying, it is

$$\begin{gathered} r{x}^4-1=0 \\ {x}^4=1 \\ x=1 \end{gathered}$$

This derivative $f^{\text{'}}$ is zero at the point$x=1$and always exists. Applying the first derivative test and testing signs at both ends of the point$x=1,$it is

$$\begin{gathered} f^{\text{'}}\left(0\right)=0^4-1 \\ =-1 \end{gathered}$$

And

$$\begin{gathered} f^{\text{'}}\left(2\right)=2^4-1 \\ =15 \end{gathered}$$

Thus sign chart is shown below

Thus by the first derivative test,  has a local minimum at

The function has no local maxima.

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

Since the sign of  measures the concavity of you can find inflection points by looking for the places where changes sign.

Here

Now

Testing for sign, it is

And

Thus the function has inflection point at .

Thus the graph of  is shown below