Limits, Continuity & Differentiability
Increasing/Decreasing Functions
Grade 12
Question:
<p>We have <i>f</i>(<i>x</i>) = <i>e</i><sup><i>x</i></sup> − <i>x</i> and <i>g</i>(<i>x</i>) = <i>x</i><sup>2</sup> − <i>x</i>. If <i>f</i>(<i>g</i>(<i>x</i>)) is an increasing function, then <i>x</i> belongs to</p>
<p>\(\left[0, \frac{1}{2}\right] \cup [1, \infty)\)</p>
<p>\(\left(0, \frac{1}{2}\right) \cup (1, \infty)\)</p>
<p>\([0, \infty)\)</p>
<p>\((-\infty, 0] \cup \left[\frac{1}{2}, 1\right]\)</p>
Step-by-Step Solution
Key Concept: A composite function f(g(x)) is increasing when the derivative f'(g(x))·g'(x) > 0, requiring both factors to have the same sign. Since f'(u) = e^u - 1 is always positive for u > 0, we need g(x) > 0 and g'(x) > 0 simultaneously.
<p><strong>Step 1:</strong> Find the derivative using chain rule: (f∘g)'(x) = f'(g(x))·g'(x)</p><p><strong>Step 2:</strong> Calculate f'(u) = e^u - 1 and g'(x) = 2x - 1</p><p><strong>Step 3:</strong> So (f∘g)'(x) = (e^(g(x)) - 1)(2x - 1)</p><p><strong>Step 4:</strong> For f∘g to be increasing: (e^(g(x)) - 1)(2x - 1) > 0</p><p><strong>Step 5:</strong> Since e^(g(x)) - 1 > 0 requires g(x) > 0, we need: x² - x > 0 AND 2x - 1 > 0</p><p><strong>Step 6:</strong> From x² - x > 0: x(x - 1) > 0 ⟹ x < 0 or x > 1</p><p><strong>Step 7:</strong> From 2x - 1 > 0: x > 1/2</p><p><strong>Step 8:</strong> Taking intersection: x > 1 (since we need both conditions)</p><p>∴ Answer: x ∈ (1, ∞)</p>
Correct Answer: A