# Antiderivative

#### rushin25

Find f. (x>0) f ''(x) = x^-2
x > 0
f(1) = 0
f(5) = 0

i am having trouble finding f(x) because the denominator goes to infinity... Last edited:

#### Random Variable

$$\displaystyle f'(x) = \frac{x^{2}}{2} - 2x + C$$

$$\displaystyle f(x) = \frac{x^{3}}{6} - x^{2} + Cx + D$$

$$\displaystyle f(1) = 0 = \frac{1}{6} -1 + C + D$$

$$\displaystyle f(5) = 0 = \frac{125}{6} - 25 + 5C + D$$

now just solve the above two equations simultaneously to find the values of C and D

• HallsofIvy

#### rushin25

i m sorry .... i guess i made a mistake while posting the question...it is

f "(x)= x^-2
x > 0
f(1) = 0
f(5) = 0

find f(x)

#### Random Variable

i m sorry .... i guess i made a mistake while posting the question...it is

f "(x)= x^-2
x > 0
f(1) = 0
f(5) = 0

find f(x)
$$\displaystyle f'(x) = -\frac{1}{x} + C$$

$$\displaystyle f(x) = - \ln x + Cx + D$$

$$\displaystyle f(1) = 0 = -\ln 1 + C + D = C + D$$

$$\displaystyle f(5) = 0 = -\ln 5 + 5C + D$$

so $$\displaystyle C = \frac{\ln 5}{4}$$

and $$\displaystyle D = - \frac{\ln 5}{4}$$

#### HallsofIvy

MHF Helper
i m sorry .... i guess i made a mistake while posting the question...it is

f "(x)= x^-2
x > 0
f(1) = 0
f(5) = 0

find f(x)
It goes to infinity at x= 0 but that has nothing to do with this problem. It says explicitely (in your first post) that you are to find f(x) for x> 0.