Chapter 4.6, Problem 40E

To determine

To calculate: At which point on the curve $y=1+40x^3-3x^5$ does the tangent line have the largest slope.

Answer to Problem 40E

The two points are $\left(2,225\right)\text{and }\left(-2,-223\right)$ .

Explanation of Solution

Given information:

A point on the curve $y=1+40x^3-x^5$

Formula used:

Area of right angled triangle: $=\frac{1}{2}\times \text{base}\times \text{perpendicular}$

Point-Slope form of the equation of a line is

  $y-y_1=m\left(x-x_1\right)$

Where $m$ is the slope and $\left(x_1,y_1\right)$ is the point in the line.

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

A point on the curve $y=1+40x^3-x^5$

The curve

  $y=1+40x^3-3x^5\mathrm{......}\left(1\right)$

For slope, differentiate with respect to $x$

  $y^{\prime }=120x^2-15x^4$

Let us denote the slope of tangent as $M\left(x\right)$

  $M\left(x\right)=120x^2-15x^4$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate with respect to $x$

  $M^{\prime }\left(x\right)=240x-60x^3\mathrm{......}\left(2\right)$

Solve for $M^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}240x-60x^3=0\\ \Rightarrow 60x\left(4-x^2\right)=0\\ \Rightarrow x=0,4-x^2=0\\ \Rightarrow x=0,x=2,x=-2\end{array}$

Differentiate equation $\left(2\right)$ with respect to $x$

  $M^{″}\left(x\right)=240-180x^2$

At $x=0$

  $M^{″}\left(0\right)=240$

At $x=2$

  $\begin{array}{l}{M}^{″}\left(2\right)=240-180\times 2^2\\ =-480\end{array}$

At $x=-2$

  $\begin{array}{l}{M}^{″}\left(-2\right)=240-180\times {\left(-2\right)}^2\\ =-480\end{array}$

Therefore,

  $M^{″}\left(x\right)$ is negative at $x=\pm 2$ , hence there is local maximum at $x=\pm 2$

Now put $x=2$ and $x=-2$ in equation $\left(1\right)$ and solve

  $\begin{array}{l}y=1+40\times 2^3-3\times 2^5\\ =1+320-96\\ =225\end{array}$

And

  $\begin{array}{l}y=1+40\times {\left(-2\right)}^3-3{\left(-2\right)}^5\\ =1-320+96\\ =-223\end{array}$

By second derivate test gives us maximum.

Conclusion:

Thus the two points are $\left(2,225\right)\text{and }\left(-2,-223\right)$ .