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Computing Difference Quotients - Problem 1
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.