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

$f''(x) = -\dfrac1{x^2}$

$f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1$

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

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

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

$-5C_1 = \ln(6)$

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

Then

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