Applications of Derivatives
Extrema of Composite Functions
Grade 12
Question:
<p>Let <i>f</i>(<i>x</i>) = <i>x</i>² − 2<i>x</i> and <i>g</i>(<i>x</i>) = <i>f</i>(<i>f</i>(<i>x</i>) − 1) + <i>f</i>(5 − <i>f</i>(<i>x</i>)), then</p>
<p>(a) <i>g</i>(<i>x</i>) ≥ 0, for all <i>x</i> ∈ ℝ</p>
<p>(b) <i>g</i>(<i>x</i>) ≤ 0, for some <i>x</i> ∈ ℝ</p>
<p>(c) <i>g</i>(<i>x</i>) > 0, for some <i>x</i> ∈ ℝ</p>
<p>(d) <i>g</i>(<i>x</i>) < 0, for all <i>x</i> ∈ ℝ</p>
Step-by-Step Solution
Key Concept: We need to express g(x) in terms of f(x), then analyze the behavior of f and substitute strategically. Since f(x) = x² - 2x = (x-1)² - 1, we can use this form to simplify the composition and determine the sign of g(x).
<p><strong>Step 1:</strong> Simplify f(x). We have f(x) = x² - 2x = (x-1)² - 1. Note that f(x) ≥ -1 for all x ∈ ℝ, with minimum value -1 at x = 1.</p><p><strong>Step 2:</strong> Let u = f(x). Then we need to find g(x) = f(u-1) + f(5-u).</p><p><strong>Step 3:</strong> Compute f(u-1). We have f(u-1) = (u-1)² - 2(u-1) = u² - 2u + 1 - 2u + 2 = u² - 4u + 3.</p><p><strong>Step 4:</strong> Compute f(5-u). We have f(5-u) = (5-u)² - 2(5-u) = 25 - 10u + u² - 10 + 2u = u² - 8u + 15.</p><p><strong>Step 5:</strong> Add them together: g(x) = f(u-1) + f(5-u) = (u² - 4u + 3) + (u² - 8u + 15) = 2u² - 12u + 18 = 2(u² - 6u + 9) = 2(u-3)².</p><p><strong>Step 6:</strong> Since g(x) = 2(u-3)² = 2(f(x)-3)², and since (f(x)-3)² ≥ 0 for all values of f(x), we have g(x) ≥ 0 for all x ∈ ℝ.</p><p><strong>Step 7:</strong> Verify: g(x) = 0 when f(x) = 3, i.e., when x² - 2x = 3, which gives (x-3)(x+1) = 0, so x = 3 or x = -1. At these points g achieves its minimum value of 0.</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a