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

Question

f ''(x) = x⁻²,    x > 0,    f(1) = 0,    f(6) = 0

Answer 1

Integrating:

f'(x) = -x^(-1)+ C

F(x) = -ln(x) + Cx+k

F(1) = 0 ---> -ln(1) + C(1) + k = 0

0 + C + k = 0

C+k = 0

F(6) = 0 ---> -ln(6) + 6C + k = 0

-ln(6) + 6C + k = 0

C+k = 0

-ln(6) + 6C + k = 0

subtracting:

5C - ln(6) = 0

5C = ln(6)

C = ln(6)/5 ---> k = -ln(6)/5

F(x) = -ln(x) + Cx+k

= -ln(x) + ln(6)/5 * X - ln(6)/5

= -ln(x) + ln(6)/5 [ x-1]

check:

F'(x) = -1/x + ln(6)/5

F''(x) = Dx[-x^(-1)] = x^(-2)

F(1) = -ln(1) + ln(6)/5 * (1-1) = 0 + ln(6)*0 = 0

F(6) = -ln(6) + ln(6)/5 * (6-1) = -ln(6) + ln(6)/5*(5) = -ln(6) + ln(6) = 0

F(x) = -ln(x) + ln(6)/5 [ x-1]

Proven.