Can someone help me with this problem? Find f if f''(x) = x^-2, x>0, f(1)=0, and f(2)=0.

I tried using antiderivatives but I realized I couldn't take the antiderivative of x^-1.

There IS an antiderivative (integral) of 1/x. It is the natural log-base-e (ln) function.

In your case

f'(x) = -1/x + C1

f"(x) = - ln x + C1 x + C2

where C1 and C2 are arbitrary constants

To solve this problem, we need to find the function f(x), given that its second derivative is x^-2, while satisfying the conditions f(1) = 0 and f(2) = 0.

Let's start by integrating the given second derivative, f''(x) = x^-2.

Taking the integral of both sides, we get:

∫f''(x) dx = ∫x^-2 dx.

Integrating f''(x) gives us f'(x) + C1, where C1 is the constant of integration.

Integrating x^-2 gives us -x^-1 + C2, where C2 is another constant of integration.

Therefore, our equation becomes:

f'(x) + C1 = -x^-1 + C2.

Next, we want to find f(x) by integrating once more. Integrating both sides of the equation with respect to x, we get:

∫(f'(x) + C1) dx = ∫(-x^-1 + C2) dx.

Integrating f'(x) with respect to x gives us f(x) + C1x + C3, where C3 is another constant of integration.

Integrating -x^-1 with respect to x gives us -ln(x) + C4, where C4 is yet another constant of integration.

So now our equation becomes:

f(x) + C1x + C3 = -ln(x) + C4.

To find the specific values of C1, C3, C4, we can use the initial conditions f(1) = 0 and f(2) = 0.

For f(1) = 0:

0 + C1(1) + C3 = -ln(1) + C4.
C1 + C3 = C4.

For f(2) = 0:

0 + C1(2) + C3 = -ln(2) + C4.
2C1 + C3 = C4 - ln(2).

Solving both of these equations, we can find the specific values of C1, C3, C4.

Once we have those values, we substitute them back into the equation:

f(x) + C1x + C3 = -ln(x) + C4.

This will give us the specific function f(x) that satisfies f''(x) = x^-2 and the given initial conditions.