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Problem 28 Hard Difficulty

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.

$ f(x) = \dfrac{x^2 - 1}{2x - 3} $

Answer

Domain of $f:\left(-\infty, \frac{3}{2}\right) U\left(\frac{3}{2}, \infty\right) ;$ Domain of $f^{\prime}:\left(-\infty, \frac{3}{2}\right) U\left(\frac{3}{2}, \infty\right)$

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Video Transcript

you are trying to find the derivative of the function F of X equals x squared minus 1/2, X minus three. And in doing so, we want to use the definition of the derivative and the definition of the derivative is f prime of X equals the limit As Delta X approaches zero of F of X plus delta x minus F of X. All over delta X. So that means we have to start by evaluating the function at X plus delta X. So F prime of X. We'll leak with a limit As Delta X is approaching zero of X plus delta X squared minus 1/2 times the quantity X plus delta x minus three. And then we have to subtract. So we'll subtract the original function which is x squared minus 1/2, X minus three. All over delta X. Now, before we could subtract those two fractions in the numerator, we would either need a common denominator or get rid of the denominator altogether. So what I'm gonna do is I'm going to multiply by a giant form of one and that one is going to look like both denominators. So we're going to have two times the quantity X plus delta x minus three. And we're going to have to X -3. So when I share that whole thing to this first fraction, what will happen is this denominator will cancel with that part and I will be left with X plus delta X, quantity squared minus one, multiplied by the two X minus three. And then when I share that whole thing to this second fraction The denominator here will cancel with the two X -3. And I'll be left with mhm X squared minus one times two times x plus delta X -3. Yeah and in the denominator I will have the delta X. Which was down in the denominator to begin with. And both of those factors Yeah. Now don't forget we do need that limit as delta X is approaching zero in the front. That is very critical to be carried through step after step. So the next thing we do is I'm going to square this final meal and I'm going to distribute here and I'm going to distribute here. So my next step I'll have the limit As Delta X is approaching zero of X squared plus two X delta X plus delta x squared minus one, Multiplied by two x -3 minus the binomial X squared minus one times two X plus two delta x minus three. And my denominator will have three factors delta X two X plus two delta x minus three And two X -3. So now I've got to do some major distributing In order to clean this up. So I'm gonna start by sharing the two X. To all the parts. So I'll get limit as delta X is approaching zero. Uh two x cubed plus four x square delta x plus two X delta x squared minus two X. Then I'll distribute the -3. So I'll have negative three X squared minus six X delta x minus three delta X squared plus three. Yeah I'll then have a subtraction sign. And over here I'm now going to distribute the X squared so I'll have two X cubed plus two X squared delta x minus three X squared. Mhm I'm also going to distribute the -1. So I'll have minus two X minus two delta X plus three. And it'll be all over those three factors I want to do some combining like terms and cleaning things up a little bit here. So the first thing I'm going to do at this stage is I'm going to distribute this negative throughout the entire quantity. So we will have The limit as Delta X is approaching zero of two X cubed plus four X squared delta X plus two X delta x squared minus two X minus three X squared minus six X delta x minus three. Delta X squared plus three minus two X cubed minus two, X squared delta X plus three X squared plus two X plus two delta x minus three. All over delta X times two X plus two. Delta x minus three times the quantity of two x minus three. In doing so you're going to find a lot of light terms that are going to cancel each other out. So we'll have a negative two X cubed will cancel with a positive two X cubed a negative two X will cancel the positive two X And a positive three will cancel with a -3. You'll also have some like terms for X squared delta X and negative two X squared delta X can combine. So we will now have the limit. As Delta X approaches zero of two X square delta X plus two X delta X squared. I just noticed another set. This negative three X squared will cancel with this positive three X squared. So we still have a negative six X delta X. And negative three delta X squared. And we have a positive to delta X remaining in the numerator. And in the denominator we still have our three factors. Now in the numerator, if you closely look at that you will see that there is a greatest common factor. So you can now factor out a delta X. Leaving you with two X squared plus two X delta x minus six X minus three. Delta X plus two Over your three factors. And in turn the delta X in the numerator will cancel with the delta X. And the denominator. And at this point you can now do direct substitution and substitute zero in for all of the delta X. Is. So when you substitute your zero in we will end up with two X squared plus two X times zero minus six X minus three times zero plus two. And in the denominator you'll have two X plus two times zero minus three times the quantity of two x minus three. So these terms are now out of play leaving you with two X squared minus six X plus two over the quantity of two x minus three times two x minus three. And if you want to you can factor out a greatest common factor in the numerator and be left with two times the quantity X squared minus three X plus one. And since both factors in the denominator are identical, we could Display it as two X -3, the quantity squared, so that is equivalent to our derivative.