f''(x) = x^-2, x>0, f(1)=0, f(7)=0. Find f.

Answer 1

assuming f(1)=0 vs. f'(1)=0

f'(x)=-(1/x)+C₁

f(x)=-ln(x)+C₁x+C₂

initial conditions for f

0=C₁+C₂

0=-ln7+7C₁+C₂

C₁=(ln7)/6

C₂=-(ln7)/6

f(x)=[(ln7)/6)](x-1)-lnx

Answer 2

f"(x) = x^-2 = 1/x^2

f'(x) = x^-1/-1+c= -1/x +c

f'(1)=0=-1+1

f'(x)=-1/x+1

f(x)= -lnx+x+c2

f(7)=0=-ln7 +7 +c2

f(x) = -lnx +x +ln7-7