Find and simplify the difference quotient calculator

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The Difference Quotient formula is $$ {f(x+h)-f(x)\over h} $$ which computes the red secant line slope - that connects the 2 given points - below.



The Difference Quotient can be looked at in 2 different ways.


1) In Algebra we learn that the slope m of a line passing through the two points (x1,y1) and (x2,y2) is \[ m= {y2-y1 \over x2-x1} \]
2) Moving on to Calculus, we generalize this concept to Functions. We compute the so called Secant Line Slope as the slope through the two points (x,f(x)) and (x+h,f(x+h)) that lie on the Function f(x).

Thus, the formula for the Difference Quotient is \[ {f(x+h)-f(x) \over (x+h)-x } \] which simplifies to the above Difference Quotient Formula \[ {f(x+h)-f(x)\over h} \] .

The slope of the red line in the above image is the secant line slope which is exactly what the Difference Quotient finds.

Example1 (simple): If \( f(x)=2x+5 \) then its Difference Quotient is \[ {f(x+h)-f(x) \over h } ={2(x+h)+5 - (2x+5) \over h} = {2h\over h} = 2 \] which is the slope between any 2 points on the line 2x+5.
That was not complicated ;-)

Example2 (more advanced): If \( f(x)=x^2+5x \) then its Difference Quotient is \[ {f(x+h)-f(x) \over h} = {(x+h)^2+5(x+h) - (x^2+(5x)) \over h} = \] \[ {(x^2+2xh+h^2)+5x+5h - (x^2+5x) \over h} = {2xh+h^2+5h \over h} = 2x+h \]
That was a bit more complicated. Now try the above Difference Quotient Calculator a few times.

The Difference Quotient in Calculus:

In Calculus, we extend the idea of a Difference Quotient. It is used to find the slope at one point instead of between two points. To accomplish this, we take the limit as h->0 of (f(x+h)-f(x))/h which moves the "dummy" point (x+h, f(x+h)) towards point (x,f(x)).
By finding the slope between those two approaching points we actually find the slope at (x,f(x)). This is called "Finding the Slope at a Point by using the Limit of a Difference Quotient of a Function" in Calculus, quite a mouthful.

Example: If \( f(x)=x^2 \) then \[ {f(x+h)-f(x) \over h} = {(x+h)^2 - x^2 \over h } = {2hx +h^2\over h} = 2x + h = 2x\] as limit h->0

This tells us that the slope at any point on the graph of f(x)=x^2 is 2x thanks to the Difference Quotient and taking the Limit h->0.

For instance, the slope of f(x)=x^2 at x=5 is computed as 2*5 = 10.

Here is another Difference Quotient calculator that you may like: https://calculator-online.net/difference-quotient-calculator/

You may also like this How-to-Find-A-Difference-Quotient Video at https://www.youtube.com/watch?v=eyn1ARLkLcA

Here is how: The Difference Quotient computes the Average Rate of Change between 2 given points.
Example: The Average Rate of Change of f(x)=x^2 over the interval [1,3] is (f(3)-f(1))/(3-1) = (9-1)/(2) = 4 .

Since x=2 lies between 1 and 3 we can use their Average Rate of Change, 4, as an approximation to the Instantaneous Rate of Change at x=2.

The exact Instantaneous Rate of Change is found by computing the derivative of x^2 , which is 2x , and evaluating it at x=2 yielding also 4.

This perfect match between the instantaneous rate of change and average rate is actually a result of the Mean Value Theorem in Calculus. For details about the Mean Value Theorem read Paul's Online Notes.

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