suppose ( f(-1) = 5 ), ( f(-1) = 2 ), and ( g(x) = x^{3}f(x) ).\nthe tangent line to ( g(x) ) at ( x = -1 )…

Explanation:

Step1: Find (g(-1))

Substitute (x = - 1) into (g(x)=x^{3}f(x)). (g(-1)=(-1)^{3}f(-1)) Since (f(-1) = 5), then (g(-1)=(-1)\times5=-5)

Step2: Find (g^{\prime}(x)) using the product rule

The product rule states that if (y = u(x)v(x)), then (y^{\prime}=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)). Let (u(x)=x^{3}) and (v(x)=f(x)). Then (u^{\prime}(x) = 3x^{2}) and (v^{\prime}(x)=f^{\prime}(x)) So (g^{\prime}(x)=3x^{2}f(x)+x^{3}f^{\prime}(x))

Step3: Find (g^{\prime}(-1))

Substitute (x=-1), (f(-1) = 5) and (f^{\prime}(-1)=2) into (g^{\prime}(x)) (g^{\prime}(-1)=3(-1)^{2}f(-1)+(-1)^{3}f^{\prime}(-1)) (g^{\prime}(-1)=3\times1\times5+(-1)\times2) (g^{\prime}(-1)=15 - 2=13)

Step4: Use the point - slope form of a line

The point - slope form of a line is (y - y_{0}=m(x - x_{0})), where ((x_{0},y_{0})) is a point on the line and (m) is the slope. Here (x_{0}=-1), (y_{0}=g(-1)=-5) and (m = g^{\prime}(-1)=13) (y-(-5)=13(x - (-1))) (y + 5=13(x + 1)) Expand: (y+5=13x+13) (y=13x + 8)

Answer:

(y = 13x+8)