Fall $2024$

$2.8.$ Suppose $f(x)$ is a function such that $f(-3)=2$ and $f^{\text{'}}(-3)=-4.$ Let $g(x)=2x^{2}f(3x).$

$($a$)$ Find an expression for $g^{\text{'}}(x).$

$g(x)=2x^2*f(3x)$

$g^{\text{'}}(x)=(4x)(f(3x))+(2x^2)(3f)$

$=(2x^2)(-3(3x))$

$=-18x^3$

$2=-1\mathrm{.}8x^3$

$\sqrt[3]{-\frac{1}{9}}=\sqrt[3]{x^3},x=\sqrt[3]{-\frac{1}{9}}$

$($b$)$ Given that $g^{\text{'}}(-1)=-32,$ find an equation of the tangent line to $g(x)$ at $x=-1.$ Note: You can use the given information to check your answer in part $($a$).$

$\backslash$table$[[$equation of tangent line:,$y+\frac{1}{9}=-32(x+1)$