Question

Find f. f ″(x) = x^−2, x > 0, f(1) = 0, f(6) = 0

Answer 1

If you do in fact mean
`f(1)=f(6)=0` (as opposed to one of these being the derivative of
`f` at some point), then integrating twice gives

`f''(x) = -\frac1{x^2}`

`f'(x) = \displaystyle -\int (dx)/(x^2) = \frac1x + C_1`

`f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2`

From the initial conditions, we find

`f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0`

`f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)`

Eliminating
`C_2`, we get

`(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))`

`-5C_1 = \ln(6)`

`C_1 = -\frac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)`

Then

`\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}`