question 7 of 14, step 1 of 1 given g(x) = √(3x⁴ + 7x²), use the derivative rule for inverse functions to…

Explanation:

Step1: Recall the formula for the derivative of an inverse function

The formula for the derivative of the inverse function $(g^{-1})'(y)$ is $(g^{-1})'(y)=\frac{1}{g'(x)}$, where $y = g(x)$. We first need to find $x$ such that $g(x)=g(- 2)$ and then find $g'(x)$.

Step2: Find $g(-2)$

Substitute $x = - 2$ into $g(x)=\sqrt{3x^{4}+7x^{2}}$. [ \begin{align*} g(-2)&=\sqrt{3(-2)^{4}+7(-2)^{2}}\ &=\sqrt{3\times16 + 7\times4}\ &=\sqrt{48+28}\ &=\sqrt{76}=2\sqrt{19} \end{align*} ]

Step3: Differentiate $g(x)$ using the chain - rule

Let $u = 3x^{4}+7x^{2}$, then $g(x)=\sqrt{u}=u^{\frac{1}{2}}$. The derivative of $u$ with respect to $x$ is $u'=12x^{3}+14x$. By the chain - rule $g'(x)=\frac{1}{2\sqrt{3x^{4}+7x^{2}}}\times(12x^{3}+14x)=\frac{12x^{3}+14x}{2\sqrt{3x^{4}+7x^{2}}}=\frac{6x^{3}+7x}{\sqrt{3x^{4}+7x^{2}}}$.

Step4: Evaluate $g'(-2)$

Substitute $x=-2$ into $g'(x)$: [ \begin{align*} g'(-2)&=\frac{6(-2)^{3}+7(-2)}{\sqrt{3(-2)^{4}+7(-2)^{2}}}\ &=\frac{6\times(-8)-14}{\sqrt{48 + 28}}\ &=\frac{-48-14}{\sqrt{76}}\ &=\frac{-62}{2\sqrt{19}}=-\frac{31}{\sqrt{19}} \end{align*} ]

Step5: Use the inverse - derivative formula

Since $(g^{-1})'(g(-2))=\frac{1}{g'(-2)}$, substituting the value of $g'(-2)$ we found above: [ (g^{-1})'(g(-2))=-\frac{\sqrt{19}}{31} ]

Answer:

$-\frac{\sqrt{19}}{31}$