Computing Difference Quotients - Problem 2
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.