2015 AP Calculus BC 5a | AP Calculus BC solved exams | AP Calculus BC | Khan Academy

Consider the function ( f(x) ) is equal to ( \frac{1}{x^2} - Kx ) where ( K ) is a nonzero constant. The derivative of ( f ) is given by, and they give us this expression right over here. It's nice that they took the derivative for us.

Now part A, let ( K ) equal 3 so that ( f(x) ) is equal to ( \frac{1}{x^2} - 3x ). So they said ( K ) equal to three. Write an equation for the line tangent to the graph of ( F ) at the point whose x-coordinate is four.

To find an equation for a line, the equation of a line is going to be of the form ( y = mx + b ) where ( m ) is the slope of the line and ( b ) is the y-intercept. The slope of the line right over here, this needs to be equal to the derivative evaluated when ( x ) is equal to 4.

So we could say ( y = ) or let me write it this way, we could say that ( m ) is going to be equal to ( F' ) when ( x ) is equal to 4. So ( F' ) of 4 which is equal to, well we know that ( K ) is equal to three. They gave us ( F' ) of ( x ), so it's going to be ( 3 - \frac{2 \cdot 4}{4^2 - 3 \cdot 4} ) squared.

Now, this is an eight right over here. All I did is ( F' ) of ( x ) when ( K ) is equal to 3 is going to be ( 3 - \frac{2x}{x^2 - 3x} ), and all of that squared. I want to evaluate what ( F' ) of four is. So every place where I saw an ( x ), I substitute it with a four. Where I saw the ( k ), ( k ) is three, and so this is going to be equal to the numerator ( 3 - 8 ) is (-5) over, this is ( 16 - 12 ) which is going to be ( 4 ).

So ( 16 - 12 ) is ( 4 ), and then we square it, so it's going to be ( \frac{-5}{4} ) squared. And so let me write this way: ( m = \frac{-5}{16} ).

So how do we figure out ( b )? Now, what are the coordinates when ( x ) is equal to 4? What is ( y ) going to be equal to? Well, ( Y = f(x) ), so we know that ( y ) on the curve, we know that ( Y ) is going to be equal to ( f(4) ), so before we evaluated ( f' ) of four, now we're going to evaluate ( y ) as being ( f(4) ), which is equal to ( \frac{1}{4^2} - 3 \cdot 4 ).

That is equal to ( \frac{1}{16 - 12} ) which is ( \frac{1}{4} ). So this point right here when ( x ) is 4, then ( y ) is equal to ( \frac{1}{4} ).

So we can use that information to solve for ( b ) when ( y ) is ( \frac{1}{4} ). So we're going to say ( y = m \frac{-5}{16} x + b ). Well, when ( y = \frac{1}{4} ) and ( x = 4 ), then plus ( b ).

So I can now solve for ( b ). All I did is I used ( F' ) of ( x ) to figure out ( m ) when ( x ) is equal to 4. Then I said, okay, well what is the value of ( y ) when ( x ) is equal to 4? So if I know ( y ), ( m ), and ( x ), then I can solve for ( b ).

So let's just do that: ( \frac{1}{4} = 4 \cdot \frac{-5}{16} + b ). I can add ( \frac{5}{4} ) to both sides, and I get ( \frac{5}{4} + \frac{1}{4} = b ) or ( b = \frac{6}{4} ) which you could say, well there's a bunch of ways you could write this.

We could just say this is equal to ( 1.5 ). So our equation is ( y = \frac{-5}{16} x + 1.5 ) or if we wanted to write everything as a fraction, we could say ( y = \frac{-5}{16} x + \frac{3}{2} ).

And there you go.