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Problem 30 Hard Difficulty

Evaluate the difference quotient for the given function. Simplify your answer.

$ f(x) = \dfrac{x + 3}{x + 1} $ , $ \dfrac{f(x) - f(1)}{x - 1} $



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Video Transcript

Okay, we're going to evaluate this difference quotient for the function F of X equals X plus three over X plus one. Notice that inside that quotient, we're going to need to know f of one. So let's go ahead and find that now FF one would be one plus 3/1 plus one, so that would be four over to. And that would be too. Now, let's go ahead and substitute each of these quantities into our quotient. So f of X is X plus three over X plus one minus half of one, which is to over X minus one. We want to simplify this. We would call this a complex fraction where we have fractions inside fractions. And the best thing to do at this point is to find a special form of one to multiply by so that we can simplify and not have so many fractions. So what I would multiply by is X plus one over X plus one, and I put them over one just so everything lines up nicely. So now we're going to multiply the first fraction by X plus one and the X plus ones cancel. So we just have X plus three minus. Now we're going to multiply the second fraction by X plus one and are the second number two by X plus one. And there's nothing to cancel. So it's just two times a quantity X plus one and then on the bottom. We just have X minus one times X plus one. There's nothing to cancel their either, so we'll just write that as X minus one times X plus one. Now let's keep going and simplify and see where we get. Okay, so we'll simplify the numerator by distributing the negative to. And we have X Plus three minus two X minus two, and we're going to leave our denominator factored for the time being. Okay, now, let's combine like terms. We have X and we have minus two X. So those combined to B minus X and we have three minus two. So those combined to B plus one. So is there a relationship between negative X plus one an X minus one. It seems to me that those are opposites. He's here are opposites. So what if we factor a negative one out of the numerator? We have negative one times a quantity X minus one. And then we can cancel that with the X minus one in the denominator. So then what we have left is negative one over X plus one.