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# Computing Difference Quotients - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let’s simplify another difference quotient. This time, let’s do it for f(x) equals 9 over x minus 2. So we start with the difference quotient formula, f(x+h) minus f(x) over h.

Now for this function, f(x+h), is 9 over x plus h minus 2. 9 over x plus h minus 2. We have minus f(x) which is just this, 9 over x minus 2 and all of that over h.

I’ve got a complex fraction here and the way I usually eliminate these little denominators here is that I have a trick. I multiply the numerator and denominator of the big fraction by the least common denominator of the little fractions and that’s going to be x plus h minus 2, x minus 2. So I multiply the top and bottom by that.

In the next step there’ll be a lot of cancelling. 9 over x plus h minus 2 times all this. The x plus h minus 2's will cancel and you’ll get 9 times x minus 2 minus. Now this term times all this, the x minus 2's will cancel and you’ll get 9 times x plus h minus 2. In the denominator you’re not going to get any cancellation you’ll just have a h times these two guys so h times x plus h minus times x minus 2. Resist the temptation to multiply all this out. Turns out difference quotients are used in calculus and it’s actually nicer if you leave this factored in calculus.

Let’s simplify the numerator. We’ve got 9x minus 18 for the left-hand term, and minus 9x minus 9h plus 18 on the right-hand term, all over h times x plus h minus 2, times x minus 2. It is at this point that we look for cancellation. The 9x's cancel, the 18's cancel. So we have minus 9h over h times x plus h minus 2 times x minus 2.

And finally we can cancel the h's. And so we’re left with negative 9 over x plus h minus 2 times x minus 2. That’s our final answer.

Remember this trick; whenever you’re trying to find the difference quotient, for a rational function like this, the trick of multiplying the top and bottom by the least common denominator of the little fractions. That will clear your fractions in one step.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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