Let f be a differentiable function with f(-4)=4, f'(-4)=-3
f(-8)=-4 and f'(-8)=-5. let the function g(x)=f(2x) write the equation of the line tangent to the graph of g at the point where x=-4.

y + 10x + 44 =0

Step-by-step explanation:

Given g(x)=f(2x)

Differentiate it on both side with respect to x.

=-5*2 =-10

Slope =-10

Here x coordinate = -4

g(-4)=f(-8)=-4 = y coordinate

Thus point of tangency on function g is (-4,-4)

Thus equation of tangent is

y + 4= -10(x+4)

y + 10x + 44 =0

answer: add the !

step-by-step explanation: then we will answer