Question

$f^{\text{'}}\left(x\right)=e^x(x+4)$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$f$.

$f^{\text{'}}\left(x\right)=e^x(x+4)$

Short answer

$f$ has a local minimum at $x=-4$

The function has no local maxima

The function has inflection point at$x=-5$

Step-by-step solution

Given information

The given derivative of a function is

$f^{\text{'}}\left(x\right)=e^x(x+4)$

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 $l$.

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

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

(c) If $f^{\prime}$ is zero in the interior of $l$, then $f$ is constant on $J$.

Suppose both $f$ and $f^{\prime}$ are differentiable on an interval $/$, then If $f^{\prime \prime}$ is positive on $I$, then $f$ is concave up on $I$.

Again

If $f^{t}$ 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 $f^{\prime}(x)$ 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 $f^{\prime}(x)$ 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 $f^{\prime}(x)$ 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 $f^{\prime}(x)$ is negative for both $x \in(a, c)$ and $x \in(c, b)$, then $f$ does not have a local cxtremum at $x=c$.

Here

$$

f^{\prime}(x)=e^{x}(x+4)

$$

Now simplifying, it is

$$

\begin{aligned}

e^{x}(x+4) &=0 \\

x+4 &=0 \\

x &=-4

\end{aligned}

$$

This derivative $f^{\prime}$ is zero at the point $x=-4$ and always exists. Applying the first derivative test and testing signs at both ends of the point $x=-4$, it is

$$

\begin{aligned}

f^{\prime}(-5) &=e^{-5}(-5+4) \\

&=e^{-5}(-1) \\

&=-6.7

\end{aligned}

$$

And

$$

\begin{aligned}

f^{\prime}(0) &=e^{0}(0+4) \\

&=4

\end{aligned}

$$

Thus sign chart is shown below

Thus by the first derivative test, $f$ has a local minimum at $x=-4$.

The function has no local maxima.

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

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

Here

$$

\begin{aligned}

f^{\prime \prime}(x) &=\frac{d}{d x}\left(e^{x}(x+4)\right) \\

&=\left(\frac{d}{d x} e^{x}\right)(x+4)+e^{x}\left(\frac{d}{d x}(x+4)\right) \\

&=e^{x}(x+4)+e^{x}

\end{aligned}

$$

Now

$$

\begin{aligned}

e^{x}(x+4)+e^{x} &=0 \\

e^{x}(x+4) &=-e^{x} \\

x+4 &=-\frac{e^{x}}{e^{x}} \\

x+4 &=-1 \\

x=-5

\end{aligned}

$$

Testing for sign, it is

$$

\begin{aligned}

f^{\prime \prime}(-6) &=e^{-6}(-6+4)+e^{-6} \\

&=-0.002

\end{aligned}

$$

And

$$

\begin{aligned}

f^{\prime \prime}(0) &=e^{0}(0+4)+e^{0} \\

&=5

\end{aligned}

$$

Thus the function has inflection point at $x=-5$

Thus the graph of $f$ is shown below