Applications of Derivatives
Extrema of Functions
Grade 12
Question:
<p>Let \(f(x) = \tan^{-1}\left(\frac{1-x}{1+x}\right)\). Then the difference of the greatest and least value of \(f(x)\) on \([0,1]\) is:</p>
<p>(a) \(\frac{\pi}{2}\)</p>
<p>(b) \(\frac{\pi}{4}\)</p>
<p>(c) \(\pi\)</p>
<p>(d) \(\frac{\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the identity $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$ to simplify.
<p>Note that $f(x) = \tan^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}(1) - \tan^{-1}(x) = \frac{\pi}{4} - \tan^{-1}(x)$. At $x=0$, $f(0) = \frac{\pi}{4}$. At $x=1$, $f(1) = 0$. The difference is $\frac{\pi}{4} - 0 = \frac{\pi}{4}$.</p>
Correct Answer: b