Question - Finding Tangent Line to Composition of Functions
Solution:
Given $$ g(x) = \frac{16}{x} $$ and $$ f(x) = -4 \sqrt{x} $$, find the tangent line to $$ g(f(x)) $$ at $$ x=4 $$.Step 1: Compute $$ f(4) $$.$$ f(4) = -4 \sqrt{4} = -4 \cdot 2 = -8 $$Step 2: Compute $$ g(f(4)) $$.$$ g(-8) = \frac{16}{-8} = -2 $$Step 3: Compute $$ f'(x) $$.$$ f'(x) = -4 \cdot \frac{1}{2}x^{-\frac{1}{2}} = -2x^{-\frac{1}{2}} $$Step 4: Compute $$ f'(4) $$.$$ f'(4) = -2 \cdot 4^{-\frac{1}{2}} = -2 \cdot \frac{1}{2} = -1 $$Step 5: Compute $$ g'(x) $$.$$ g'(x) = -16x^{-2} $$Step 6: Compute $$ g'(-8) $$.$$ g'(-8) = -16(-8)^{-2} = -16 \cdot \frac{1}{64} = -\frac{1}{4} $$Step 7: Apply chain rule to find derivative of $$ g(f(x)) $$ at $$ x=4 $$.$$ (g(f(x)))' = g'(f(x)) \cdot f'(x) $$$$ (g(f(4)))' = g'(-8) \cdot f'(4) = -\frac{1}{4} \cdot (-1) = \frac{1}{4} $$Step 8: Write the equation of the tangent line.$$ y = mx + b $$$$ m = \frac{1}{4} $$$$ y = \frac{1}{4}x + b $$Step 9: Use point $$ (f(4), g(f(4))) = (4, -2) $$ to solve for $$ b $$.$$ -2 = \frac{1}{4} \cdot 4 + b $$$$ -2 = 1 + b $$$$ b = -3 $$Step 10: Final equation of the tangent line.$$ y = \frac{1}{4}x - 3 $$
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