Relations & Functions
Inverse functions
Grade 12
Question:
<p><strong>137.</strong> If \(g(x)\) and \(h(x)\) are invertible functions and \(h(x)=3g(x)+7\), then \(h^{-1}(x)\) is equal to:</p>
<p>(a) \(3g^{-1}(x)-7\)</p>
<p>(b) \(\dfrac{1}{3g^{-1}(x)+7}\)</p>
<p>(c) \(\dfrac{1}{3}g^{-1}(x)+7\)</p>
<p>(d) \(g^{-1}\!\left(\dfrac{x-7}{3}\right)\)</p>
Step-by-Step Solution
Key Concept: Since h(x) = 3g(x) + 7, find the inverse by reversing operations: subtract 7, then divide by 3, then apply g⁻¹. The inverse function undoes operations in reverse order of how they were applied.
<p><strong>Step 1:</strong> Start with h(x) = 3g(x) + 7. Let y = h(x), so y = 3g(x) + 7.</p><p><strong>Step 2:</strong> Solve for g(x) in terms of y: y - 7 = 3g(x), so g(x) = (y - 7)/3.</p><p><strong>Step 3:</strong> Apply g⁻¹ to both sides: g⁻¹(g(x)) = g⁻¹((y - 7)/3), which gives x = g⁻¹((y - 7)/3).</p><p><strong>Step 4:</strong> Replace y with x to get the inverse function: h⁻¹(x) = g⁻¹((x - 7)/3).</p><p><strong>Verification:</strong> h(h⁻¹(x)) = h(g⁻¹((x-7)/3)) = 3g(g⁻¹((x-7)/3)) + 7 = 3·(x-7)/3 + 7 = x ✓</p><p>∴ Answer: h⁻¹(x) = g⁻¹((x - 7)/3)</p>
Correct Answer: D