Chapter 5, Problem 5RE

a. $$

To determine

Tofindthe interval on which the function $y=1+x-x^2-x^4$ is increasing.

Answer to Problem 5RE

The function is then increasing for $(-\infty ,0.385]$ .

Explanation of Solution

Given information:

The given function is $y=1+x-x^2-x^4$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=1+x-x^2-x^4$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=1-2x-4x^3$

  $y^{″}=-2-12x^2$

Here the second derivative is always negative, so the function is concave down for all values of $x$ .

  $\begin{array}{l}{y}^{\prime }=1-2x-4x^3=0\\ x\approx 0.38545\dots \end{array}$

Here, the function is positive until one point which is at zero and then negative.

Therefore, the function is increasing for $(-\infty ,0.385]$ .

b. $$

To determine

To find the interval on which the function $y=1+x-x^2-x^4$ is decreasing.

Answer to Problem 5RE

The function is then decreasing for $x>0.385$ .

Explanation of Solution

Given information:

The given function is $y=1+x-x^2-x^4$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=1+x-x^2-x^4$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=1-2x-4x^3$

  $y^{″}=-2-12x^2$

Here the second derivative is always negative, so the function is concave down for all values of $x$ .

  $\begin{array}{l}{y}^{\prime }=1-2x-4x^3=0\\ x\approx 0.38545\dots \end{array}$

Here, the function is positive until one point which is at zero and then negative.

Therefore, the function is decreasing for $x>0.385$ .

c. $$

To determine

To find the interval on which the function $y=1+x-x^2-x^4$ is concave up.

Answer to Problem 5RE

The function is never concave up.

Explanation of Solution

Given information:

The given function is $y=1+x-x^2-x^4$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=1+x-x^2-x^4$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=1-2x-4x^3$

  $y^{″}=-2-12x^2$

Here the second derivative is always negative, so the function is concave down for all values of $x$ and there are no inflection points.

  $\begin{array}{l}{y}^{″}=-2-12x^2=0\\ -12x^2=2\\ x^2=-\frac{1}{6}\end{array}$

This has no solution so there are no values of $x$ that makes $y^{″}=0$ .

  $-2-12x^2<0\forall x$

Therefore, the function is never concave up.

d. $$

To determine

To find the interval on which the function $y=1+x-x^2-x^4$ is concave down.

Answer to Problem 5RE

The function is always concave down.

Explanation of Solution

Given information:

The given function is $y=1+x-x^2-x^4$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=1+x-x^2-x^4$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=1-2x-4x^3$

  $y^{″}=-2-12x^2$

Here the second derivative is always negative, so the function is concave down for all values of $x$ and there are no inflection points.

  $\begin{array}{l}{y}^{″}=-2-12x^2=0\\ -12x^2=2\\ x^2=-\frac{1}{6}\end{array}$

This has no solution so there are no values of $x$ that makes $y^{″}=0$ .

  $-2-12x^2<0\forall x$

Therefore, the function is always concave down.

e. $$

To determine

To find the interval on which the function $y=1+x-x^2-x^4$ has local extreme values.

Answer to Problem 5RE

The function haslocal maximum at $\left(0.385,1.215\right)$ .

Explanation of Solution

Given information:

The given function is $y=1+x-x^2-x^4$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=1+x-x^2-x^4$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=1-2x-4x^3$

Equating the equation to zero,

  $\begin{array}{l}{y}^{\prime }=1-2x-4x^3=0\\ x\approx 0.385\end{array}$

Now to determine if the critical value $x\approx 0.385$ is a maximum or minimum,

  $\begin{array}{l}{y}^{\prime }\left(0\right)=1-2\left(0\right)-4{\left(0\right)}^3=1>0\\ y^{\prime }\left(1\right)=1-2\left(1\right)-4{\left(1\right)}^3=-5<0\\ \end{array}$

Since the derivative switches from positive to negative, the critical point is a local maximum. It is also an absolute maximum since there are no endpoints or other critical points.

  $y\left(0.385\right)=1+0.385-{0.375}^2-{0.385}^4\approx 1.215$

Therefore, the function has local maximum at $\left(0.385,1.215\right)$ .

f. $$

To determine

To find the interval on which the function $y=1+x-x^2-x^4$ has inflection points.

Answer to Problem 5RE

The function has noinflection point.

Explanation of Solution

Given information:

The given function is $y=1+x-x^2-x^4$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=1+x-x^2-x^4$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=1-2x-4x^3$

  $y^{″}=-2-12x^2$

The point of inflection occurs at the values $x$ that make the second derivative undefined or equal to zero.

  $\begin{array}{l}{y}^{″}=-2-12x^2=0\\ -12x^2=2\\ x^2=-\frac{1}{6}\\ x=\pm \sqrt{-\frac{1}{6}}\end{array}$

Since the square root of a negative number is not a real number, there are no values of $x$ that make $y^{″}=0$ .

Since there are no values of $x$ that make $y^{″}=0$ or be undefined, the second derivative is strictly positive or strictly negative.

  $\begin{array}{l}{y}^{″}\left(0\right)=-2-12{\left(0\right)}^2=-2<0\\ -12x^2=2\\ x^2=-\frac{1}{6}\end{array}$

  $y^{″}=-2-12x^2$ is always negative, so the function is concave down for all values of $x$ and there are no inflection points.

Therefore, the function has no inflection point.