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## Differential Calculus (2017 edition)

### Course: Differential Calculus (2017 edition)>Unit 11

Lesson 1: Linear approximation

# Linear approximation of a rational function

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.F (LO)
,
CHA‑3.F.1 (EK)
Sal finds a linear expression that approximates y=1/(x-1) around x=-1. This is done by finding the equation of the line tangent to the graph at x=-1, a process called "linear approximation.".

## Want to join the conversation?

• What's the benefit of using linear approximation in this specific function? It's as easy/hard to calculate any point of the graph with the original function as it is with the linear approximation (which is only less accurate). So why use it here?
• With this particular function, it's easy enough to plug in whole numbers, but it might be more difficult to plug in decimal numbers that don't have nice fraction equivalents, especially if you don't have a calculator.
• Could anyone explain (or tell me the video explaining) why (y - y_1) = m(x - x_1) ?
I'm only familiar with (y = mx + b) version and I can't even come up with keywords for search.
• those two formulas are the two basic forms of a line
y=mx+b (standard)
y-y1=m(x-x1) (point-slope)

However, those two equations are equivalent, let's see.

( y - y1 ) = m ( x - x1 )
( y - y1 ) = mx - mx1
y = mx - mx1 + y1
y = mx + ( y1 - mx1 ) = mx + b
which means b= y1 - mx1 , this is the formula calculating y- intercept of the line at any point with the slope of the line.

And that's it.
• is this the sames as the taylor series?
• You are correct -- linear approximation is equivalent to using a first degree Taylor series.
• How do you calculate the Error and Percentage Error after finding the Linear Approximation? (if you could add a video on it, it would be greatly appreciated)
• %Error = (Approximate Value - Exact Value) / Exact Value
• At time stamp , Sal is calculating the value of the linear approximation using the point slope formula in the form, (y-y1)/(x-x1)=b, and he points to b and calls it the slope. But I always thought that b was the y intercept. So b would be equal to: (y-y1) – m(x-x1)=b, and that would be the y intercept, not the slope. Anyway, I'm sure he wanted to calculate the slope here, so it seems to me that he should have written (y-y1) / (x-x1) = m, not b. Am I missing something?
• What grade math is this?
• The beginning of calculus is usually 12th grade math and/or early college math. Some advanced math students take calculus earlier than 12th grade.

Have a blessed, wonderful day!
• I'm still not sure what the difference between this and finding the equation of the tangent line is. Could someone try to help me understand this?
• They're the same thing. A linear approximation to a function at a point is the same as its tangent line at that point.
• At Sal after applying the power rule says "then we're going to multiply that times the derivative of x minus 1 with respect to x."

I have been following all Calculus lectures and exercises so far and this step is a little confusing for me, could someone please help me?
• So he is just multiplying 'that' with the derivative of x-1 (with respect to x) which is 1.
Hope this helps! If you have any questions or need help, please ask! :)