wws: problem 15\n(1 point)\nlet (f(x)), (g(x)), and (h(x)) be differentiable functions with the following…

wws: problem 15\n(1 point)\nlet (f(x)), (g(x)), and (h(x)) be differentiable functions with the following properties:\n- (f(x)=8x^{2}+3).\n- the equation of the tangent line of (g(x)) at (- 1) is (y = 4x-10), and\n- (h = fcirc g).\nfind (h(-1)).\nremark: recall that the symbol (circ) denotes composition, so (h = fcirc g) means (h(x)=f(g(x))).\npreview my answers submit answers\nyou have attempted this problem 1 time.\nyour overall recorded score is 0%.\nyou have unlimited attempts remaining.

wws: problem 15\n(1 point)\nlet (f(x)), (g(x)), and (h(x)) be differentiable functions with the following properties:\n- (f(x)=8x^{2}+3).\n- the equation of the tangent line of (g(x)) at (- 1) is (y = 4x-10), and\n- (h = fcirc g).\nfind (h(-1)).\nremark: recall that the symbol (circ) denotes composition, so (h = fcirc g) means (h(x)=f(g(x))).\npreview my answers submit answers\nyou have attempted this problem 1 time.\nyour overall recorded score is 0%.\nyou have unlimited attempts remaining.

Answer

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)