Finding specific antiderivatives: rational function | AP Calculus AB | Khan Academy

So we're told that ( F(2) ) is equal to 12. ( F' ) prime of ( x ) is equal to ( \frac{24}{x^3} ), and what we want to figure out is what ( F(-1) ) is.

Alright, so they give us the derivative in terms of ( x ), so maybe we can take the antiderivative of the derivative to find our original function. So let's do that. We could say that ( F(x) ) is going to be equal to the antiderivative, or we could say the indefinite integral of ( F' ) prime of ( x ), which is equal to ( \frac{24}{x^3} ).

I could write it over like this ( \frac{24}{x^3} ), but to help me process it a little bit more, I'm going to write this as ( 24x^{-3} ) because then it'll become a little clearer how to take that antiderivative ( \frac{d}{dx} ).

So what is the antiderivative of ( 24x^{-3} )? Well, we're just going to do the power rule in reverse. So what we're going to do is we're going to increase the exponent. Let me just rewrite it: it's going to be ( 24x^{-\frac{3}{1}} ), we're going to increase the exponent by 1, so it's going to be ( x^{-3 + 1} ) and then we're going to divide by that increased exponent, which is ( -3 + 1 ).

So that is going to be ( -3 + 1 = -2 ), and then we divide by ( -2 ). And if you're in doubt about what we just did, we're kind of doing the power rule in reverse now. Take the power rule, take the derivative of this using the power rule: ( -2 \times 24 = -48 \div -2 ) is just going to be 24, and then you decrement that exponent going to ( -3 ).

So are we done here? Is this ( F(x) )? Well, ( F(x) ) might involve a constant, so let's put a constant out here because notice if you were to take the derivative of this thing here, the derivative of ( \frac{24x^{-2}}{-2} ) we already established is ( 24x^{-3} ), but then if you take the derivative of a constant, well that just disappears, so you don't see it when you look at the derivative.

So we have to make sure that there might be a constant. And I have a feeling, based on the information that they've given us, that we're going to make use of that constant. So let me rewrite ( F(x) ). So we know that ( F(x) ) can be expressed as ( -12x^{-2} + C ).

So how do we figure out that constant? Well, they have told us what ( F(2) ) is. ( F(2) ) is equal to 12, so let's write this down. So when ( F(2) = 12 ), which is equal to ( -12(2^{-2}) + C ).

So ( 12 = -12(2^{-2}) + C ). Now, what is this ( 2^{-2} )? ( 2^{-2} = \frac{1}{2^2} = \frac{1}{4} ). So this is ( -12 \times \frac{1}{4} = -3 ).

So it's ( -3 + C ). Now we can add 3 to both sides to solve for ( C ). We get ( 15 = C ), so ( C = 15 ).

That is equal to 15. And so now we can write our ( F(x) ) as ( F(x) = -12x^{-2} + 15 ). And now using that, we can evaluate ( F(-1) ).

( F(-1) ): wherever we see an ( x ), we put in ( -1 ). So this is going to be ( -12(-1)^{-2} + 15 ).

So ( F(-1) = -12 \div (-1)^{-2} + 15 ). Well, ( (-1)^{-2} ) is just 1, so it's going to be ( -12 + 15 ), which is equal to 3. And we're done! This thing is equal to 3.