Science:Math Exam Resources/Courses/MATH110/April 2011/Question 08 (e)

MATH110 April 2011

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Question 08 (e)
Let $\displaystyle f(x)=\ln \left(4-x^{2}\right)$ Determine where ƒ is concave up and where it is concave down.

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Hint
A function's concavity is related to the sign of its second derivative.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We find the second derivative of $f$ by using the quotient rule on the first derivative: ${\begin{aligned}f''(x)&={\frac {2(x^{2}-4)-2x(2x)}{(x^{2}-4)^{2}}}\\&={\frac {-2x^{2}-8}{(x^{2}-4)^{2}}}\end{aligned}}$ We find possible inflection points by setting $\displaystyle f''(x)$ equal to zero and also by determining where it is undefined. In this case, $\displaystyle f''$ is never zero and is undefined when $x=-2,2$. Because these points are not included in our domain, ƒ has no inflection points, i.e. ƒ does not change its concavity. Therefore, it suffices to find the concavity at any point in the domain, e.g. at x = 0: $f''(0)={\frac {-8}{16}}<0$ and hence ƒ is concave down on its whole domain.

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