Chapter 4.3, Problem 8E

To determine

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

Answer to Problem 8E

It has been determined that the graph of the given function is concave up in the interval $\left(0,2\right)$ and concave down in the intervals $\left(-\infty ,0\right)$ and $\left(2,\infty \right)$ .

Explanation of Solution

Given:

The function, $y=-x^4+4x^3-4x+1$ .

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 $y=-x^4+4x^3-4x+1$ .

Differentiating,

  $y^{\prime }=-4x^3+12x^2-4$

Differentiating again,

  $y^{″}=-12x^2+24x$

Simplifying,

  $y^{″}=-12x\left(x-2\right)$

Let the second derivative be positive.

Then, $y^{″}=-12x\left(x-2\right)>0$ .

This implies that either $x<0$ and $x>2$ OR $x>0$ and $x<2$ .

Now, it is not possible for $x<0$ and $x>2$ to hold true simultaneously.

So, the only possibility is $x>0$ and $x<2$ .

Combining, $0<x<2$ .

Now, let the second derivative be negative.

Then, $y^{″}=-12x\left(x-2\right)<0$ .

This implies that either $x<0$ and $x<2$ OR $x>0$ and $x>2$ .

Combining, it follows that either $x<0$ OR $x>2$ .

So, the second derivative is positive when $0<x<2$ and the second derivative is negative when $x<0$ and when $x>2$ .

Equivalently, the second derivative is positive in the interval $\left(0,2\right)$ and the second derivative is negative in the intervals $\left(-\infty ,0\right)$ and $\left(2,\infty \right)$ .

According to the given criteria, it follows that the graph of the given function is concave up in the interval $\left(0,2\right)$ and concave down in the intervals $\left(-\infty ,0\right)$ and $\left(2,\infty \right)$ .

Conclusion:

It has been determined that the graph of the given function is concave up in the interval $\left(0,2\right)$ and concave down in the intervals $\left(-\infty ,0\right)$ and $\left(2,\infty \right)$ .