Question

wws: problem 15
(1 point)
let (f(x)), (g(x)), and (h(x)) be differentiable functions with the following properties:

• (f(x)=8x^{2}+3).
• the equation of the tangent line of (g(x)) at (- 1) is (y = 4x-10), and
• (h = fcirc g).

find (h(-1)).
remark: recall that the symbol (circ) denotes composition, so (h = fcirc g) means (h(x)=f(g(x))).
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Explanation:

Step1: Find the derivative of \(f(x)\)

Given \(f(x)=5x^{2}+3\), then \(f'(x)=\frac{d}{dx}(5x^{2}+3)=10x\)

Step2: Find \(g(- 1)\) and \(g'(-1)\)

The equation of the tangent - line of \(y = g(x)\) at \(x=-1\) is \(y = 4x - 10\).
The point - slope form of a tangent line is \(y - y_0=g'(x_0)(x - x_0)\). When \(x=-1\), \(y = 4\times(-1)-10=-14\). So \(g(-1)=-14\) and \(g'(-1) = 4\)

Step3: Use the chain - rule

The chain - rule states that if \(h(x)=f(g(x))\), then \(h'(x)=f'(g(x))\cdot g'(x)\)
We want to find \(h'(-1)\), so we substitute \(x = - 1\) into the chain - rule formula:
\(h'(-1)=f'(g(-1))\cdot g'(-1)\)
Since \(g(-1)=-14\) and \(g'(-1)=4\), and \(f'(x)=10x\), then \(f'(g(-1))=f'(-14)=10\times(-14)=-140\)
\(h'(-1)=f'(g(-1))\cdot g'(-1)=-140\times4=-560\)

Answer:

\(-560\)