Question

Let f be a differentiable function with f(-3)= 1 and f'(−3)= 1 and f'(-3)= -5
Let the function g(x)=3[f(x)]^2
Write the equation of the line tangent to the graph of gg at the point where x= -3

Answer 1

First, how can you say that "f'(−3)= 1 and f'(-3)= -5" ?? Pick one. I pick 1
using the chain rule, we have g' = 6f f'
g'(-3) = 6f(-3) f'(-3) = 6(1)(-1) = -6
g(3) = 3(-3)^2 = -27
So now we have a point and a slope, giving us the equation
g+27 = -6(x+3)