Question

In Exercises \(7-12\) , use the Concavity Test to determine the intervals on which the graph of the function is (a) concave up and (b) concave down. $$y=-x^{4}+4 x^{3}-4 x+1$$

Short answer

The function is concave up on the intervals \(-\infty to \frac{1}{3}\) and \(1 to \infty\) and concave down on the interval \(\frac{1}{3} to 1\).

Step-by-step solution

Find the Derivative

To start off, differentiate the function \(y=-x^{4}+4 x^{3}-4 x+1\). The first derivative \(y'\) is given by this formula: \(y'=f'(x)=4x^{3}-12x^{2}+4\).

Find the Second Derivative

Next, differentiate \(f'(x)\) to get the second derivative \(f''(x)\). This will give us \(f''(x)=12x^{2}-24x +4\).

Determine where the second derivative is zero

To find where the graph changes concavity, we solve \(f''(x)=0\) for \(x\). That is, \(12x^{2}-24x +4=0\). Divide by 4, we have \(3x^{2}-6x +1=0\). Solving this quadratic equation using the quadratic formula \(x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) where \(a=3\), \(b=-6\), and \(c=1\), we get \(x=1\) and \(x=\frac{1}{3}\).

Test Intervals

We now have two critical points, and they divide the real line into three intervals: \(-\infty to \frac{1}{3}\), \(\frac{1}{3} to 1\), and \(1 to \infty\). We choose test points in each interval and evaluate \(f''\) at these points. If \(f''\) is positive, the function is concave up on that interval. If it's negative, the function is concave down. For \(-\infty < x < \frac{1}{3}\) (test with \(x=0\)), \(f''(x)>0\). So the function is concave up on this interval. Similarly, for \(\frac{1}{3} < x < 1\) (test with \(x=\frac{1}{2}\)), \(f''(x)<0\). So the function is concave down on this interval. Lastly, for \(1 < x < \infty\) (test with \(x=2\)), \(f''(x)>0\). So the function is concave up on this interval.