Applications of Derivatives
Finding Maximum and Minimum Values
Grade 12
Question:
<p>The extremum values of the function <span class="latex">f(x) = \frac{1}{\sin x + 4} + \frac{1}{\cos x + 4}</span>, where <span class="latex">x \in \mathbb{R}</span> are</p>
<p>(a) <span class="latex">\frac{4}{8 + 2\sqrt{2}}</span></p>
<p>(b) <span class="latex">\frac{2}{8 + 2\sqrt{2}}</span></p>
<p>(c) <span class="latex">\frac{4}{4\sqrt{2} + 1}</span></p>
<p>(d) <span class="latex">\frac{2}{8 + \sqrt{2}}</span></p>
Step-by-Step Solution
Key Concept: Find critical points by setting the derivative to zero and evaluate the function at those points
<p><strong>Solution:</strong></p><p>Given, <span class="latex">f(x) = \frac{1}{\sin x + 4} + \frac{1}{\cos x + 4}</span></p><p><span class="latex">f'(x) = \frac{-\cos x}{(\sin x + 4)^2} + \frac{-\sin x}{(\cos x + 4)^2}</span></p><p>For extremum, set <span class="latex">f'(x) = 0</span>:</p><p><span class="latex">\frac{\cos x}{(\sin x + 4)^2} = \frac{\sin x}{(\cos x + 4)^2}</span></p><p>At critical point where <span class="latex">\sin x = \cos x</span> (i.e., <span class="latex">x = \frac{\pi}{4}</span>):</p><p><span class="latex">f\left(\frac{\pi}{4}\right) = \frac{2}{\frac{1}{\sqrt{2}} + 4} = \frac{4}{8 + 2\sqrt{2}}</span></p>
Correct Answer: a