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Computing Difference Quotients - Problem 1 6,672 views

Teacher/Instructor 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 a difference quotient. This is a difference quotient; f(x+h) minus f(x) over h and we’ll simplify it for the function f(x) equals 5x² minus 3x. Let’s start by writing down our difference quotient and then we’ll replace f(x+h) with this function, replacing x with (x+h), 5(x+h)² minus 3 times (x+h). So that’s f(x+h) minus, I need more line here, f(x) which is just 5x² minus 3x all over h.

I need to expand this binominal. I get 5 times x² plus 2xh plus h², minus 3x, minus 3h, minus, and distribute this minus sign over these two terms. Minus 5x², minus minus, plus 3x all over h. And then I distribute the 5 over this trinomial. 5x² plus 10xh plus 5h² minus 3x. minus 3h. minus 5x² minus 3x, all of that over h. A lot of stuff cancels, this is the fun part.

5x² and 5x² cancels. 3x and 3x cancels. Anything else? No cancellation in the numerator. We have 10xh plus 5h² minus 3h. I’ll write that over here. Equals 10xh plus 5h² minus 3h all over h. And you’ll notice each of these terms has a factor of h. We can factor that h out and then cancel. So we have 10x plus 5h minus 3 times h, all over h. You can cancel the h and you’re left with 10x plus 5h minus 3. And that’s the simplified difference quotient. That’s your answer.

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