Science:Math Exam Resources/Courses/MATH100/December 2010/Question 04 (d)/Solution 1

To determine the intervals of concavity, we must evaluate ƒ''(x) and determine where it is positive and negative:

$f(x)={\begin{cases}{\frac {4}{\pi }}\tan ^{-1}(x)&{\text{if }}x\geq 1\\2-x^{4}&{\text{if }}x<1\end{cases}}$

Evaluating the second derivative of f gives

$f''(x)={\begin{cases}{\frac {4}{\pi }}{\frac {-2x}{(1+x^{2})^{2}}}&{\text{if }}x>1\\-12x^{2}&{\text{if }}x<1\end{cases}}$

From this we see that the possible inflection points are x = 0. We need only confirm that the sign of the concavity changes across x = 0. We must also consider concavity changes across x = 1 since ƒ is not differentiable there.

When x < 1, the second derivative is always negative since -12x² is always negative.

When x > 1, the second derivative is always negative as well since in that case, -2x is always negative.

From these results we can see that ƒ(x) is concave down for all x except at the critical points. In other words, ƒ(x) is concave down on $(-\infty ,0),(0,1),(1,\infty )$ and so, there are no inflection points.