tangent lines with derivative rules\nquestion\n1. let ( f ) be a differentiable function with ( f(2)= - 4 )…

tangent lines with derivative rules\nquestion\n1. 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 ).\nanswer\nanswer 1 out of 1\nsubmit answer

tangent lines with derivative rules\nquestion\n1. 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 ).\nanswer\nanswer 1 out of 1\nsubmit answer

Answer

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)