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. $$y=4 x^{3}+21 x^{2}+36 x-20$$

Short answer

The function \(y=4x^{3}+21x^{2}+36x-20\) is concave up on the interval \((-7/4, +∞ )\) and concave down on the interval \( ( -∞, -7/4 ) \).

Step-by-step solution

Find the first derivative of the function

Differentiate the given function \(y=4x^{3}+21x^{2}+36x-20\) with respect to \(x\) to obtain the first derivative. The derivative of \(x^{n}\) is \(nx^{n-1}\). Hence, the first derivative of the function \(y'\) is \(12x^{2}+42x+36\).

Find the second derivative of the function

Now differentiate the first derivative obtained above to get the second derivative. This gives \(y''=24x+42\). The second derivative of a function provides valuable information about the concavity of the original function.

Find the critical points

The solutions to the equation \(y''=0\), referred to as critical points, are the values of \(x\) where the function might change its concavity. Equating \(y''\) to zero gives \(24x+42=0\). Solving for \(x\), we get \(x=-42/24=-7/4 \).

Test intervals

Next, determine the concavity in each interval divided by the critical point \(x=-7/4\). Take any point in the interval and substitute it in the second derivative function. (a) For \(x<-7/4 = -1.75\), take \(x=-2\) in \(y''\). This gives \(y''(x=-2)=24*(-2)+42=-6 <0\), hence this interval is concave down.(b) For \(x>-7/4 = -1.75\), take \(x=0\) in \(y''\). This gives \(y''(x=0)=24*(0)+42=42 >0\), hence this interval is concave up. Thus, the function \(y\) is concave up on \(( -7/4, +∞ )\) and concave down on \( ( -∞, -7/4 ) \).

Conclusion

The intervals of concavity can indicate the overall behavior of the function within those intervals. Our function is concave up for \(x > -7/4\) and concave down for \(x < -7/4\).