Question

Multiple Choice Which of the following functions has exactly two local extrema on its domain? (A) \(f(x)=|x-2|\) (B) \(f(x)=x^{3}-6 x+5\) (C) \(f(x)=x^{3}+6 x-5\) (D) \(f(x)=\tan x\) (E) \(f(x)=x+\ln x\)

Short answer

The function \(f(x)=x^{3}-6 x+5\) has exactly two local extrema on its domain. Thus, the answer is (B) \(f(x)=x^{3}-6 x+5\).

Step-by-step solution

Analyze (A) \(f(x)=|x-2|\)

Plot the function to visualize it and find the local extrema. From the analysis, it’s clear that \(f(x)=|x-2|\) does not have any local extrema. The extremum that is present is not a local minimum or maximum, but a global one. So, choice (A) is incorrect.

Analyze (B) \(f(x)=x^{3}-6 x+5\)

Find the derivative of \(f(x)=x^{3}-6 x+5\), i.e., \(f'(x)=3x^{2}-6\). Setting it equal to zero gives \(x= \pm \sqrt{2}\). These are the critical points. Now, using the second derivative test, we find \(f''(x)=6x\). Evaluating \(f''(\sqrt{2}) > 0\) and \(f''(-\sqrt{2}) > 0\), this indicates that there are minimums at both points, which means there are exactly two local extrema. So, choice (B) is correct.

Analyze (C) \(f(x)=x^{3}+6 x-5\)

Although one should stop at (B) being correct as per the nature of multiple choice questions, for the sake of completeness, we shall analyze all options. On the same line as step 2, differentiate the function and find the critical points. The derivative is \(f'(x)=3x^{2}+6\). Unfortunately, this doesn't equate to zero for any real number, so there are no critical points, and hence no local maximum or minimum. Therefore, choice (C) is incorrect.

Analyze (D) \(f(x)=\tan x\)

Here the function \(f(x)=\tan x\) fluctuates between \(-\pi/2\) and \(+\pi/2\). The function has vertical asymptotes or undefined points at \((2n+1)\pi/2\) (where n is an integer), and it takes all values between \(-\infty\) and \(+\infty\). It has no absolute maxima or minima and it has an infinite number of local maxima and minima. Therefore, choice (D) is incorrect.

Analyze (E) \(f(x)=x+\ln x\)

Here we differentiate \(f(x)\) to get \(f'(x)=1+1/x\). This derivative exists for all \(x>0\) and it never equals zero. So there are no local maxima or minima. Therefore, choice (E) is incorrect.