Asked by Bill

Find the equation of the tangent to the graph y=f'(x) at x=3 where f(x)=x^4

So I did :
f'(x)=4x^3
y-81=4x^3(x-3)
y=4x^4-12x+81

?? Am i even remotely correct?
Answered by TutorCat
plug in 4 for x
f'(x)=4(4)^3 =?
Answered by TutorCat
I mean plug in 3 for x
f'(x)=4(3)^3 =?
Answered by Bill
108
Answered by TutorCat
y-81=108(x-3)
and solve for y.
Answered by Bill
y=108x-243
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