No brainlets

I don't know how to do this either. Can you walk me through the proofs senpai?

Just integrate it retard.

Give me a break. I didn't get that far in uni 1. Shits advanced math yo.

Calculus is basically like higher algebra. All of high school mathematics leads up to this.

You could probably learn this shit at khan academy.

11 = 3c1 + c2 + (256/51)*3^(17/8)

What the fuck?

This board is full of brainlets.

B-but you end up with two unknowns in one linear equation.

You fucked up.

Okay, what's the solution?

What is the question?

I see nothing but 2 definitions. What is there to prove?

Not being a brainlet :^)

From the statement of the derivative of f, we find a function involving two constants. Then we know the value of f for one value of x, so we get one equation with two unknowns.
??

There, I integrated it for you plebs. Now do the rest. Plug in.

Nevermind, figured it out.
You can find C1 in terms of C2 or vice versa, then have a funtion with only one constant, then use f(3) = 11 again to find that one constant.

After integration f(x)=12*-2/5*x^(-5/2)+C=(-24/5)x^(-5/2).
f(3)=(-24/5)*3^(-5/2)+C=11
C=11+(24/5)*3^(-5/2)
So
f(x)=(-24/5)x^(-5/2)+11+(24/5)*3^(-5/2)

3.2*x^(-1.5) + 10.38

Your first line is just f' though, not f.
I don't see how to get around the two constants.
Doing the integral to find f' yields an expression with one constant. Then integrating to find f yields an expression with a new constant, so you end up with an expression with two constants.

Like, its -4.8*x^(-2.5) + 11.30792014356

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God diggity damn calculus is amazing.