Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Let \(k\) and \(K\) be the minimum and the maximum values of the function \(f(x) = \dfrac{(1+x)^{0.6}}{1+x^{0.6}}\) in \([0, 1]\), respectively, then the ordered pair \((k, K)\) is equal to</p>
<p>\((1, 2^{0.6})\)</p>
<p>\((2^{-0.4}, 2^{0.6})\)</p>
<p>\((2^{-0.6}, 1)\)</p>
<p>\((2^{-0.4}, 1)\)</p>
Step-by-Step Solution
Key Concept: Find critical points by taking the derivative and checking boundary values. The function's behavior depends on whether the numerator grows faster or slower than the denominator as x increases from 0 to 1.
<p><strong>Step 1:</strong> Evaluate at boundaries.</p><p>f(0) = (1+0)^0.6/(1+0) = 1/1 = 1</p><p>f(1) = (1+1)^0.6/(1+1^0.6) = 2^0.6/2</p><p><strong>Step 2:</strong> Find f'(x) using quotient rule.</p><p>f'(x) = [0.6(1+x)^(-0.4)(1+x^0.6) - (1+x)^0.6·0.6x^(-0.4)] / (1+x^0.6)²</p><p>= [0.6(1+x)^(-0.4)[(1+x^0.6) - (1+x)x^(-0.4)]] / (1+x^0.6)²</p><p><strong>Step 3:</strong> Analyze the sign of f'(x) on (0,1).</p><p>For x ∈ (0,1): (1+x)^0.6 < (1+x^0.6), so numerator of f'(x) > 0, thus f'(x) > 0.</p><p>This means f is strictly increasing on [0,1].</p><p><strong>Step 4:</strong> Identify extrema.</p><p>Minimum: k = f(0) = 1</p><p>Maximum: K = f(1) = 2^0.6/2 = 2^(-0.4)</p><p>∴ (k, K) = (1, 2^(-0.4)) or equivalently (1, 1/2^0.4)</p>
Correct Answer: D