Relations & Functions
Inverse Functions
Grade 12
Question:
<p>Let <span class="math">f:[2, ∞) → [1, ∞)</span> defined by <span class="math">f(x) = 2^{\frac{x^4 - 4x^2}{}}}</span> be an invertible function. Find <span class="math">f^{-1}(x)</span>.</p>
<p>(a) <span class="math">\sqrt{2 + \sqrt{4 - \log_2 x}}</span></p>
<p>(b) <span class="math">\sqrt{2 + \sqrt{4 + \log_2 x}}</span></p>
<p>(c) <span class="math">\sqrt{4 + \sqrt{2 + \log_2 x}}</span></p>
<p>(d) <span class="math">\sqrt{4 - \sqrt{2 + \log_2 x}}</span></p>
Step-by-Step Solution
Key Concept: To find the inverse function, swap variables and solve for the original variable. Apply logarithm to exponential equations and use the quadratic formula.
<p>To find <span class="math">f^{-1}(x)</span>, let <span class="math">y = 2^{\frac{x^4 - 4x^2}{4}}</span></p><p>Taking <span class="math">\log_2</span> on both sides: <span class="math">\log_2 y = \frac{x^4 - 4x^2}{4}</span></p><p>So <span class="math">4\log_2 y = x^4 - 4x^2</span></p><p>Rearranging: <span class="math">x^4 - 4x^2 - 4\log_2 y = 0</span></p><p>Using the quadratic formula with <span class="math">u = x^2</span>: <span class="math">u = \frac{4 ± \sqrt{16 + 16\log_2 y}}{2} = 2 ± 2\sqrt{1 + \log_2 y}</span></p><p>Since <span class="math">x ≥ 2</span>, we have <span class="math">x^2 ≥ 4</span>, so <span class="math">x^2 = 2 + 2\sqrt{1 + \log_2 y}</span></p><p>Therefore: <span class="math">f^{-1}(x) = \sqrt{2 + \sqrt{4 + \log_2 x}}</span></p><p>∴ Answer is (b).</p>
Correct Answer: B