Question

Find the tangent line approximation to
12/x
near
x = 12?

Answer 1

Can you find the 1st derivative of f(x) = 12/x?

Answer 2

-12/x^2 right?

Answer 3

That's it.

f(x) = 12/x

f'(x) = -12/x^2

Now you need to evaluate those at x = 12 and you will have all the information you need.

Answer 4

So Im I trying to find f(12)+fprime(12)(x-12)?

Answer 5

Yes.

Answer 6

But I dont get where do i get the x from?

Answer 7

Problem statement? x = 12

Answer 8

Well the equation we got in class is f(a)+fprime(a)(x-a)

Answer 9

and the teacher said that a is equal to the x your are given

Answer 10

so i get 1+(-.083(x-12)

Answer 11

1) Why would you convert it to decimals?
2) Who cares about 'a' or 'fred" or whatever. You need the equation of a line.

You have a point: (12,f(12)) = (12,1)
You have a slope: f'(12) = -1/12

That's all you need for a line if you remember the Point-Slope Form. There is just no need to memorize yet another form for "linearization". You already know how to write the equation of a line.

Answer 12

So it should be y=13/2(X-12)?

Answer 13

Well, let's see...

\(y - 1 = -\dfrac{1}{12}(x-12)\)

\(y = -\dfrac{1}{12}(x-12) + 1\)

Why do we have to write more than that?

\(12y = -(x-12) + 12\)
\(12y = -x + 12 + 12\)
\(12y = -x + 24\)
\(x + 12y = 24\)

\(\dfrac{x}{24} + \dfrac{y}{2} = 1\)

Lots of ways to write it. Which do you find suitable?

Answer 14

okay yeah thanks it was just showing up wrong because I wasn't supposed to enter it as y=

Answer 15

Perfect.