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Mlmniolc (fany) of t2 Find the !crtical funcbon_ J) = r+ 31= 10Find tbe%" +4S12=-e10Sf [har limn ju) =-M]Find all tvabcel esmpets @r)o3 34 Eaph ofthz funcbon J...

Question

Mlmniolc (fany) of t2 Find the !crtical funcbon_ J) = r+ 31= 10Find tbe%" +4S12=-e10Sf [har limn ju) =-M]Find all tvabcel esmpets @r)o3 34 Eaph ofthz funcbon Jl) = 1(1-1)F(1)= ~10,Find trc folonnns limitlim [ (x) + E()] 1 = (d 1 E }=IncaSTI nLFud & EnrFind t 1-Valuz (fet Mmlon fx) = E DOI cununuchut F1 4f(1) %$ DO" contnuous #- and f(r) {erovabl: disconbnill 2 * = Ju) Lo Condetinois 2 { Oisconinudes ae ponreniable ftx) E5 DO: contouous f(x) Bas remoiable diconnnmtyFind the slope =

mlmniolc (fany) of t2 Find the !crtical funcbon_ J) = r+ 31= 10 Find tbe %" +4S1 2=- e10 Sf [har limn ju) =-M] Find all tvabcel esmpets @r)o3 34 Eaph ofthz funcbon Jl) = 1(1-1) F(1)= ~10,Find trc folonnns limit lim [ (x) + E()] 1 = ( d 1 E }= IncaSTI nL Fud & Enr Find t 1-Valuz (fet Mmlon fx) = E DOI cununuchut F1 4 f(1) %$ DO" contnuous #- and f(r) {erovabl: disconbnill 2 * = Ju) Lo Condetinois 2 { Oisconinudes ae ponreniable ftx) E5 DO: contouous f(x) Bas remoiable diconnnmty Find the slope = of2 ln: .27281;j0 I: gapb of thz funcion Jux) -2-6r &thz pon: JG)E nocondntousM I Jx) has [etovable discontinuity 7 =22 f(x) contnuous for allilx. M=2 7 =0 ~nnt Canja%t Lbetius 3o9ntd 24



Answers

Compute $f^{\prime}(a)$ using the limits (2.1) and (2.2). $$f(x)=\frac{3}{x+1}, a=2$$

Question we solve for F prime of A given that f of X is equal to three x plus one and A is equal to one. The formula for F prime of A is as the limit of H approaches zero. We have f of A plus h minus f of a all over each. And we're gonna plug in one any time we see in a here. So plugging in will get f prime of one secret of the limit as H approaches zero of f of one plus h minus f of one all over h plenty in using the actual original F of X. Now we're gonna get f prime of one is equal to the limit As H approaches zero f of one plus h is going to be three times one plus h plus one and then to subtract f of one, we're gonna plug one into this original equation. That's three times one is three plus one. This will all be divided by H. Now we have f prime of one is equal to the limit as h approaches zero off. Let's destry that to get three plus three h plus one and distribute the negative minus three minus one all over h. So now we have f prime of one is equal to the limit as each of purchase zero Ah, we can get rid of terms that cancel each other out. So here I see we have a positive three and a negative three. We also have a positive one in a negative one. So now we have limited as each approaches to zero of three age over each. Those ages were going to cancel. Leaving us with f prime of one is equal to the limit as H approaches zero of three. And since there's no more each term, this limit basically doesn't do anything. Giving us f prime of one is equal to three.

Prime of a given that f of X is equal to three x squared plus one, and a +01 will use the form F prime of A is equal to the limit as each approaches zero of f of A plus h minus f of a divided by each. So now what we're gonna do is we're gonna plug in one, um, for a into this equation. So we have f prime of one is equal to the limit as each approaches zero of f of one plus h will be three times one plus h squared plus one minus three times one squared plus one. And that's in parentheses because we'll need to distribute that negative. We're gonna divide by H so f prime of one is equal to the limit as each approaches zero of three times, one plus h squared is going to be one plus two h plus each squared if we expand it and they will be multiplied by three. This is all plus one. And now is the time Teoh to distribute the negative sign. So this will be minus three minus one. And this is all divided by H what we can now, Dio is we can distribute the three out in front of that first term, so we'll get f prime of one is equal to three plus six h plus three h squared. Then we have plus one minus three minus one. This is all over each. So, um, I forgot my limits. Science all on that in This is the limit as h approaches zero. So now we have f prime of one is equal to the limit. As h approaches, zero of we can start to eliminate things that you know, We'll just equal zero. So we have three and negative three. We have a positive one and a negative one. So now we have the limited each purchase zero of six h plus three age squared, divided by h. We can now cancel the ages. So we have one each down here, we'll get rid of that age and this will Instead of being a three h squared, it will just go to three each. So now we have f prime of one is equal to the limit as each approaches zero of six plus three h And if we plug in zero for that h that means that we'll have f prime of one is equal to six plus zero, which is you hold a six.

We want to find prime of a given that F of X is equal to their square root of three X plus one in a is equal to one we use the formula F prime of A is equal Teoh the limit as each approaches zero off F of a plus H minus F of a, all divided by H. So in this particular instance, F prime of one is equal to the limits as each approaches zero of F of one plus h minus F of one, all divided by H. So F prime of one is equal to the limit as each approaches. Zero of F of one plus h is the square root of three times one plus h plus one. Because we've plugged in one plus h for the X value in the original function, this will be minus f of one so minus the square root of three times one plus one, all divided by each. So we get F prime of one is equal to the limit as H approaches zero of the square root of three plus three h plus one just expanding that right there, and this is minus the square root of three plus one. So just for this is all divided by H. So now to take care of this, um, all the routes in the numerator we're going to multiply by the con jacket. This is just equal to four plus three h, Soren, multiply by route four plus three h instead of minus. We're gonna do plus, so we'll do plus root for and we're gonna multiply by the same thing on the bottom Route four plus three h plus root for So now we can rewrite as f prime of one is equal to the limit as each approaches zero of four plus three h plus not plus minus four. Right, Because we have a negative and a positive those will make a negative four. This is all divided by H times square root of four plus three h and then plus root for Okay, so simplifying we get f prime of one is equal to the limit. As h approaches zero of the four and the negative four cancel out. So we get three h over h Times Route four plus three h plus root. For now, we can cancel the ages in the numerator the denominator to get F prime of one is equal to the limit as age approaches. Zero off three, divided by Route four plus three H plus root. For now, we can plug in the zero for the each value to solve it a limit so we get F prime of one is equal to three over Route four plus route for which is just equal to 3/2 plus two, so F prime of one then is equal to 3/4.

Okay, so here we have the function F of x. Y. Z is equal to E. To the X plus Y over one plus Z square. So we then take the limit as H approaches zero of f of one comma two plus H comma three minus F of +123 All over. H. That gives us the limit as H approaches zero of E. To the one plus two plus H over one plus three squared minus E to one plus 2/1 plus three squared all over each. So this is going to be equal to the limit as a church approaches zero of while we get E. Um cubed Times E to the H- E. Cubed both over 10. Um and then all over H. So as H approaches zero um while we can bring you can bring an E cube over 10. We can factor outside the limit. So this is going to be equal to e cubed over 10 and then times the limit as H approaches zero of E. To the H minus one. All over H. Okay, well as it approaches zero E. To the H is going to be equal two. Um So we this is actually gonna be equal to just um we get one here so we get one times E cube over 10, which is equal to E cube over 10 limit here is going to be equal to e cubed over 10.


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