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# Find f. f''(x) = x^{-2}, x > 0, f(1) = 0, f(4) = 0.

**Solution:**

Given: f ''(x) = x^{-2}

To find the original function, we need to integrate the given function twice.

Integrating on both the sides with respect to x.

⇒ ∫ f''(x)dx= ∫x^{-2}dx

⇒ ∫ f''(x) dx = (x^{-2 + 1})/(-2 + 1) + C\(_1\)

⇒ f'(x) = x^{-1}/(-1) + C\(_1\)

Integrating on both the sides with respect to x again

∫ f'(x) dx = -log(x) + C\(_1\)x + C\(_2\)

f(x) = -log(x) + C\(_1\)x + C\(_2\).

Given: f(1) =0

If x = 1, then f(1) = 0

0 = -log(1) + C\(_1\) + C\(_2\) --- (1)

Also f(4) = 0

0 = -log(4) + 4C\(_1\)+ C\(_2\) --- (2)

Solving (1) and (2) we get C\(_1\)_{ }and C\(_2\),

Subtracting (1) - (2)

0 = -log(1) + log(4) - 3C\(_1\)

3c_{1} = -log(1) + log(4) = log(4/1) {since log(a) - log(b) = log(a/b)}

C_{1} = log(4)/3

C_{1} = log(4)/3 and (2)

0 = -log(1) +(4/3)log(4) + C\(_2\)

log(1) - (4/3)log(4) = C\(_2\)

Replace C\(_1\) and C\(_2\) in equation in f(x)

f(x) = -log(x) + [log(4)]/3 + log(1) - (4/3)log(4)

f(x) = -log(x) + log(1) - log(4).

## Find f. f''(x) = x - 2, x > 0, f(1) = 0, f(4) = 0.

**Summary:**

If f ''(x) = x-2, x > 0, f(1) = 0, f(4) = 0 then r ⇒ f(x) = -log(x) + log(1) - log(4).

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