Question

tangent lines with derivative rules
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1. let ( f ) be a differentiable function with ( f(2)= - 4 ) and ( f(2)=3 ). let the function ( g(x)=3f(x)^{2}). write the equation of the line tangent to the graph of ( g ) at the point where ( x = 2 ).

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Explanation:

Step1: Find the derivative of \(g(x)\) using the chain - rule.

If \(g(x)=3[f(x)]^{2}\), by the chain - rule \((u^{n})^\prime = nu^{n - 1}u^\prime\), where \(u = f(x)\) and \(n = 2\). So \(g^\prime(x)=3\times2\times f(x)\times f^\prime(x)=6f(x)f^\prime(x)\).

Step2: Evaluate \(g^\prime(x)\) at \(x = 2\).

We know that \(f(2)=-4\) and \(f^\prime(2)=2\). Substitute these values into \(g^\prime(x)\): \(g^\prime(2)=6\times f(2)\times f^\prime(2)\).

Step3: Calculate the value of \(g^\prime(2)\).

\(g^\prime(2)=6\times(-4)\times2=-48\).
The equation of the tangent line to \(y = g(x)\) at \(x = a\) is given by \(y - g(a)=g^\prime(a)(x - a)\). First, find \(g(2)=3[f(2)]^{2}=3\times(-4)^{2}=3\times16 = 48\).
The equation of the tangent line at \(x = 2\) is \(y - 48=-48(x - 2)\).
Expand it: \(y-48=-48x + 96\).
Rearrange to the slope - intercept form \(y=-48x+144\).

Answer:

\(y=-48x + 144\)