Limit definition for derivation of f(x) HELPtutoring my daughter and I'm stuck

AI Thread Summary
To find the derivative of f(x) = x^2 - 4x using the limit definition, the derivative is defined as the limit of the ratio [f(x+h) - f(x)]/h as h approaches 0. By calculating f(x+h) as (x+h)^2 - 4(x+h), the expression simplifies to 2hx + h^2 - 4h. This leads to the limit of (2hx + h^2 - 4h)/h as h approaches 0. As h gets smaller, the terms involving h vanish, resulting in the derivative f'(x) = 2x - 4. Understanding this process is crucial for teaching the limit definition of derivatives effectively.
epatjn
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Homework Statement



Here's the question...use the limit defintion to find the derivation of f(x) = x^2-4x

Homework Equations



does this use the defintion of the derivative formula (using Larson, et al 4th edition of Precaclulus graphing with limits...and trying to teach someone what to do, but am at a loss at present on what to do...

The Attempt at a Solution

 
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epatjn said:

Homework Statement



Here's the question...use the limit defintion to find the derivation of f(x) = x^2-4x

Homework Equations



does this use the defintion of the derivative formula (using Larson, et al 4th edition of Precaclulus graphing with limits...and trying to teach someone what to do, but am at a loss at present on what to do...

The Attempt at a Solution

I don't have the Larson book, but I assume it defines the derivative (not _derivation_) of f at x to be the limit of the ratio [f(x+h) - f(x)]/h as h --> 0. Well, you can calculate f(x+h) and you know f(x), so you can see what the ratio is equal to. Then you can see what it becomes closer and closer to as h becomes smaller and smaller.

RGV
 


f(x)= x^2- 4x so f(x+h)= (x+h)^2- 4(x+h)= x^2+ 2hx+ h^2- 4x- 4h

f(x+h)- f(x)= x^2+ 2hx+ h^2- 4x- 4h- (x^2- 4x)= 2hx+ h^2- 4h
 
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