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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. :)

 Mar 27, 2022
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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  cool

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\\

 Mar 28, 2022
edited by Melody  Mar 28, 2022

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