A function has f’’(x) = 10 and has f’(4) = 0 and f(2) = 4. Find f(x)

f'= 10 at what x?

If f" = 10 for all x, then

f' = 10x+c
10*4+c = 0 so c = -40 and f' = 10x-40
f = 5x^2-40x+c
5*2^2-40*2+c = 4 so c=64 and f(x) = 5x^2-40x+64

ok, my eyes fail me, I can't see " vs '.

To find the function f(x), we will integrate f''(x) to get f'(x), and then integrate f'(x) to get f(x).

Given f''(x) = 10, we can integrate this equation once to find f'(x):

∫f''(x) dx = ∫10 dx
f'(x) = 10x + C1,

where C1 is the constant of integration.

Now, we have f'(4) = 0, so we can substitute x = 4 into the equation f'(x) = 10x + C1:

0 = 10(4) + C1
C1 = -40.

Therefore, f'(x) = 10x - 40.

Next, we integrate f'(x) to find f(x):

∫f'(x) dx = ∫(10x - 40) dx
f(x) = 5x^2 - 40x + C2,

where C2 is the constant of integration.

Finally, we use the given value f(2) = 4 to determine the value of C2:

4 = 5(2)^2 - 40(2) + C2
4 = 20 - 80 + C2
C2 = 64.

Hence, the function f(x) is:

f(x) = 5x^2 - 40x + 64.