**Question: **The tangent line to the graph of \(y = g(x)\) at \(x=4\) has equation \(y=-3x+11\). What is the equation of the tangent line to the graph of \(y = g(x)^3\) at \(x = 4\)?

I know that the slope is the derivative of \(g(x)\). I'm treating \(g(x)^3\) as a variable (\(g(x)^3 = x^3\)); therefore, the derivative/slope of that would be \(3x^2\), but I'm not sure where to go from there. I have an idea to find \(g(4)\) using the equation first, but I'd still appreciate some help. :)

Guest Mar 27, 2022

#1**+1 **

y=g(x)

y'=g'(x)

g'(4)=-3

The tangent of g(x) is y=-3x-11 and when x=4 y=-1 so (4,-1) is on the tangent point.

When x=4

g(4)=-1

[g(4)]^3 = (-1)^3 = -1

\(\frac{d}{dx}[g(x)]^3=3[g(x)]^2*g'(x)\\ \frac{d}{dx}[g(4)]^3=3[g(4)]^2*g'(4)\\ \frac{d}{dx}[g(4)]^3=3[-1]^2*-3\\ \frac{d}{dx}[g(4)]^3=3*1*-3\\ \frac{d}{dx}[g(4)]^3=-9\\ \)

so the gradient of the tangent of [g(x)]^3 at x=4 is -9

I did do a sample graph to check it worked at least for that example. It did

https://www.geogebra.org/classic/u7dfheht

LaTex:

\frac{d}{dx}[g(x)]^3=3[g(x)]^2*g'(x)\\

\frac{d}{dx}[g(4)]^3=3[g(4)]^2*g'(4)\\

\frac{d}{dx}[g(4)]^3=3[-1]^2*-3\\

\frac{d}{dx}[g(4)]^3=3*1*-3\\

\frac{d}{dx}[g(4)]^3=-9\\

Melody Mar 28, 2022