suppose that f'(x)= 9x for all x. compute f(2) if f(0)=0 f(5)=0 f(-3)=5initially i thought that i would have to take the integral because it works for the first one and i get the right answer which...

suppose that f'(x)= 9x for all x. compute f(2) if

f(0)=0

f(5)=0

f(-3)=5

initially i thought that i would have to take the integral because it works for the first one and i get the right answer which is 18. but there is no way possible that the integral would equal 0 at 5. even when i add "C" because C=0 when i solve for it so i have no idea how to approach the problem. Any insight/help would be greatly appreciated!

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You are on the right track. If `f'(x)=9x` then `f(x)=4.5x^2+C`

(1) If `f(0)=0` then C=0 and `f(2)=4.5(2)^2+0=18`

(2) If `f(5)=0` then `4.5(5)^2+C=0==>C=-112.5` .Then `f(2)=4.5(2)^2-112.5=18-112.5=-94.5`

(3) If `f(-3)=5` then `4.5(-3)^2+C=5==>C=5-40.5==>C=-35.5`

Then `f(2)=4.5(2)^2-35.5=-17.5`

So (a)f(2)=18  (b) f(2)=-94.5 and (c) f(2)=-17.5

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