Suppose that f'(x)=2x for all x

suppose that f'(x)=2x for all x
a)find f(-5)if f(0)=0
b)find f(-5)if f(4)=15
c)find f(-5)if f(-3)=11

well lets integrate f '(x) to find f(x).

f '(x) = 2x
f(x) = x^2 + c, where c is some unknown constant.

a)

f(0) = (0)^2 + c = 0
0 + c = 0
c = 0
f(-5) = (-5)^2 + 0 = 25.

b)

f(4) = 4^2 + c = 15
16 + c = 15
c = -1
f(-5) = (-5)^2 - 1 = 24

c)

f(-3) = (-3)^2 + c = 11
9 + c = 11
c = 2
f(-5) = (-5)^2 + 2 = 27

hello, all you would fo for this one is suppliment the number in for x. for example f(-5) = 2(-5), because two times zero is just zero so that makes sense.

you are looking for the function were f times a number equals a function
for the second one your function could be, f(x)=5x-5
where that statement would be true

and the third one could be, f(x)=x^2+2

A. F(x) is x*x so f(-5) is 25.
B. F(x) is x*x - 1 so f(-5) is 24.
C. F(x) is x*x + 2 so f(-5) is 27.

f'(x)=2x for all x => f(x) =x^2+c (constant)
a)find f(-5)if f(0)=0; x^2+c=0 => c=0, f(-5)=25
b)find f(-5)if f(4)=15; 4^2+c=15 =>c=-1, f(-5)=24
c)find f(-5)if f(-3)=11(-3)^2+c=11 =>c=2; f(-5)=27