Math
posted by Larry .
If f(x)=ax^2+b
f(3)=3 and f'(3)=2
find the coefficients a and b.

f(x) = ax^2 + b
f ' (x) = 2ax
given :
f ' (3) = 2 > 2a(3) = 2
3a = 1
a = 1/3 , so f(x) = (1/3)x^2 + b
f(3) = 3
> (1/3)(3)^2 + b = 3
3 + b = 3
b = 0
so a = 1/3, b = 0 and f(x) = (1/3)x^2
check:
f(3) = (1/3)(9) = 3
f ' (x) = (2/3)x
f '(3) = (2/3)(3) = 2
all looks good.
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