Asked by steven
Find the equation of the function f whose graph passes through the point (0, 4/3) and whose derivative is
f'(x) = x sqrt 16 − x^2
.
f(x) =
f'(x) = x sqrt 16 − x^2
.
f(x) =
Answers
Answered by
Reiny
Did you mean
f'(x) = x√(16-x^2) ??? , brackets are needed here.
then f(x) = (-1/3)(16 - x^2)^(3/2) + c
(differentiate to confirm)
if x = 4/3 , f(4/3) = 0
0 = (-1/3)(16 - 16/9)^(3/2) + c
solve for c and sub back into f(x) =
f'(x) = x√(16-x^2) ??? , brackets are needed here.
then f(x) = (-1/3)(16 - x^2)^(3/2) + c
(differentiate to confirm)
if x = 4/3 , f(4/3) = 0
0 = (-1/3)(16 - 16/9)^(3/2) + c
solve for c and sub back into f(x) =
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