Question

suppose ( f(-1) = 5 ), ( f(-1) = 2 ), and ( g(x) = x^{3}f(x) ).
the tangent line to ( g(x) ) at ( x = -1 ) is given by ( y=)

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:

\(y = 13x+8\)