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, to find h⁻¹(x), reverse the operations: subtract 7 first, then divide by 3, then apply g⁻¹. The inverse function undoes operations in reverse order.
<p><strong>Step 1:</strong> Let y = h(x) = 3g(x) + 7. To find h⁻¹(x), solve for x in terms of y.</p><p><strong>Step 2:</strong> From y = 3g(x) + 7, subtract 7: y - 7 = 3g(x)</p><p><strong>Step 3:</strong> Divide by 3: (y - 7)/3 = g(x)</p><p><strong>Step 4:</strong> Apply g⁻¹ to both sides: g⁻¹[(y - 7)/3] = x</p><p><strong>Step 5:</strong> Therefore, h⁻¹(x) = g⁻¹[(x - 7)/3]</p><p>∴ Answer: h⁻¹(x) = g⁻¹[(x - 7)/3] or equivalently g⁻¹((x - 7)/3)</p>
Correct Answer: D