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Chapter 4: Problem 66

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$f(x)=\tan ^{-1} x$$

Short Answer

Expert verified
Answer: The function is concave up for $$x < 0$$ and concave down for $$x > 0$$, with an inflection point at $$(0, 0)$$.

Step by step solution

01

Find the first derivative

To find the first derivative of the arctangent function, $$f(x)=\tan^{-1} x$$, we use the chain rule: $$f'(x) = \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$$
02

Find the second derivative

Now, we'll differentiate the first derivative one more time to get the second derivative, $$f''(x)$$: $$f''(x) = \frac{d}{dx}\left(\frac{1}{1+x^2}\right) = \frac{-2x}{(1+x^2)^2}$$
03

Identify the intervals

The function is concave up when the second derivative is positive and concave down when the second derivative is negative. We'll examine the sign of $$f''(x) = \frac{-2x}{(1+x^2)^2}$$ - For $$x > 0$$: The numerator is negative, and the denominator is always positive, so the second derivative is negative. This means that the function is concave down for $$x > 0$$. - For $$x < 0$$: The numerator is positive (since $$x$$ is negative), and again, the denominator is always positive, so the second derivative is positive. This means that the function is concave up for $$x < 0$$.
04

Find the inflection points

Inflection points occur when the second derivative changes from positive to negative or vice versa. In this case, it occurs at $$x = 0$$, so the inflection point is at $$(0, f(0)) = (0, \tan^{-1}(0)) = (0, 0)$$. So the final answer is that the function is concave up for $$x < 0$$ and concave down for $$x > 0$$, with an inflection point at $$(0, 0)$$.

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