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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.

Respuesta :

Answer:

y + 10x + 44 =0

Step-by-step explanation:

Given g(x)=f(2x)

Differentiate it on both side with respect to x.

[tex]\frac{dg(x)}{dx}=2*\frac{df(2x)}{dx}[/tex]

[tex]\frac{dg(-4)}{dx}=2*\frac{df(-8)}{dx}[/tex] =-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

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