Find f f'(x)=1+3sqrt(x) f(4)=25
1 answer:
It appears that you have a derivative and want to integrate...
dy/dx=1+3√x
y(x)=x+(2/3)x^(3/2)+C you are given that f(4)=25 so we can solve for the constant of integration...
y(4)=25=4+16/3+C
21=16/3+C
(63-16)/3=C
47/3=C so
f(x)=x+(2/3)x^(3/2)+47/3
f(x)=(3x+2x^(3/2)+47)/3
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