Question 287526: Find a function f such that the graph of f has a horizontal tangent at (2, 0) and f ''(x) = 4x.
f(x) = ?
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! Integrate (find the antiderivative) twice:
f'(x)=2x^2+C
f(x)=(2/3)x^3+Cx+D
Now, f has a horizontal tangent at (2,0) IE the function attains a local extrema and f'(2)=0
hence
f'(2)=2*2^2+C=0
implies C=-8.
Now we have f(x)=(2/3)x^3-8x+D.
But we know f(2)=0, so, (2/3)2^3-8(2)+D=0
16/3-48/3=-D
-32/3=-D
implies 32/3=D.
This gives our final function as
f(x)=(2/3)x^3-8x+32/3.
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