Chapter 4.3, Problem 7E

To determine

The intervals on which the graph of the given function is concave up and concave down.

Answer to Problem 7E

It has been determined that the graph of the given function is concave up in the interval $\left(-1.75,\infty \right)$ and concave down in the interval $\left(-\infty ,-1.75\right)$ .

Explanation of Solution

Given:

The function, $y=4x^3+21x^2+36x-20$ .

Concept used:

The graph of a function is concave up in the interval(s), where the second derivative of the function is positive and is concave down in the interval(s), where the second derivative of the function is negative.

Calculation:

The given function is $y=4x^3+21x^2+36x-20$ .

Differentiating,

  $y^{\prime }=12x^2+42x+36$

Differentiating again,

  $y^{″}=24x+42$

Let the second derivative be positive.

Then, $y^{″}=24x+42>0$ .

Simplifying,

  $x>-\frac{42}{24}$

On further simplification,

  $x>-1.75$

Now, let the second derivative be negative.

Then, $y^{″}=24x+42<0$ .

Simplifying,

  $x<-\frac{42}{24}$

On further simplification,

  $x<-1.75$

So, the second derivative is positive when $x>-1.75$ and the second derivative is negative when $x<-1.75$ .

Equivalently, the second derivative is positive in the interval $\left(-1.75,\infty \right)$ and the second derivative is negative in the interval $\left(-\infty ,-1.75\right)$ .

According to the given criteria, it follows that the graph of the given function is concave up in the interval $\left(-1.75,\infty \right)$ and concave down in the interval $\left(-\infty ,-1.75\right)$ .

Conclusion:

It has been determined that the graph of the given function is concave up in the interval $\left(-1.75,\infty \right)$ and concave down in the interval $\left(-\infty ,-1.75\right)$ .