small calculus problem

[SleepWalkerX]

There are two types of problems like these, but if I can figure out one then I can get the other.

Find k such that the line is tangent to the graph of the function:

Function:
f(x) = x^2 - kx

Line:
y = 4x - 9

answer: 2, -10

Now I assume that when it says to make the line tangent to the function, its talking about finding the derivative. But when I try to, it doesn't make sense. Would anyone care to shed some light?

 

[FelixDeCat]

And what do I get out of it?

 

[IBuyUFO]

If I look at the back of the book I too can get the answer.

 

[blueshoe]

Ok, so when you take the derivative of f(x), you end up with only the SLOPE of the tangent line and thats it.
So take the derivative of f(x):

f'(x)= 2x-k

You know that that is equal to the slope of the line, which is 4.

2x-k=4

Put that in terms of x.

x= (2+ .5k)

Plug in this x for both equations and set them equal to each other. Now you have 1 equation with 1 unknown, k.
This will be a quadtratic equation, and thats why you get those two solutions for k.

 

[thesurge]

4x-9=x^2-kx

d/dx(x^2-kx)=4

 

[TheoPetro]

find the gradiant of the function then find the equation of the line thats perpendicular to the gradient at that point. 🙂 I can overcomplicate ANYTHING!

 

[chuckywang]

You have two equations and two unknowns:

1) 2x-k = 4
2) x^2-kx = 4x-9

Combining, you get: x^2-(2x-4)*x = 4x-9, which simplifies to x^2-9 = 0, so x=3 or -3, and k=2 or -10.

 

[SleepWalkerX]

Got it!! Thanks to everyone who contributed!