Chapter 4.3, Problem 55E

To determine

The interval on which the graph of the given function is concave up.

Answer to Problem 55E

It has been determined that the graph of the given function is concave up in the interval $\left(-\infty ,-\frac{1}{a}\right)$ . So, the correct choice is option (A).

Explanation of Solution

Given:

The function, $y=a{x}^3+3x^2+4x+5$ , where $a<0$ .

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 $y=a{x}^3+3x^2+4x+5$ .

Differentiating,

  $y^{\prime }=3a{x}^2+6x+4$

Differentiating again,

  $y^{″}=6ax+6$

Now, let the second derivative be positive.

Then, $y^{″}=6ax+6>0$ .

Simplifying,

  $\begin{array}{c}6ax+6>0\\ ax+1>0\\ ax>-1\end{array}$

Since $a<0$ , it follows that,

  $x<-\frac{1}{a}$

So, the second derivative is positive when $x<-\frac{1}{a}$ .

Equivalently, the second derivative is positive in the interval $\left(-\infty ,-\frac{1}{a}\right)$ .

According to the given criteria, it follows that the graph of the given function is concave up in the interval $\left(-\infty ,-\frac{1}{a}\right)$ .

Conclusion:

It has been determined that the graph of the given function is concave up in the interval $\left(-\infty ,-\frac{1}{a}\right)$ . So, the correct choice is option (A).