4

O negatlve postlve ()" f' TRAtve 0flr) Is cancave F(x) Is increasing f(r) 1s concave down 0 flr) # cecreasing (Yi) # f""(c) nocallve than Jehtu ...

Question

O negatlve postlve ()" f' TRAtve 0flr) Is cancave F(x) Is increasing f(r) 1s concave down 0 flr) # cecreasing (Yi) # f""(c) nocallve than Jehtu Amunuaut ininimu0 f(e) = inciedeing 4 f(t) decrcaeing Tehatt Iininuint fl) then Axi) " 4 Tel HlI Rhie tue Tlcht 0' $ 1"() naralie rn1 Toura uunid TrUe Neht 0r $ Dol tlt ol f 0f (46 1* (=) Nbdmiye ch Arelsthe Vntt Knor It = () # neuslive Motatiye ertre U Dec/e

O negatlve postlve ()" f' TRAtve 0flr) Is cancave F(x) Is increasing f(r) 1s concave down 0 flr) # cecreasing (Yi) # f""(c) nocallve than Jehtu Amun uaut ininimu 0 f(e) = inciedeing 4 f(t) decrcaeing Tehatt Iininuint fl) then Axi) " 4 Tel HlI Rhie tue Tlcht 0' $ 1"() naralie rn1 Toura uunid TrUe Neht 0r $ Dol tlt ol f 0f (46 1* (=) Nbdmiye ch Arelsthe Vntt Knor It = () # neuslive Motatiye ertre U Dec/e



Answers

Table 1.12 gives values of a function $w=f(t) .$ Is this function increasing or decreasing? Is the graph of this function concave up or concave down? $$\begin{array}{c|c|c|c|c|c|c|c} \hline t & 0 & 4 & 8 & 12 & 16 & 20 & 24 \\ \hline w & 100 & 58 & 32 & 24 & 20 & 18 & 17 \\ \hline \end{array}$$

Okay, so here trying to find the intervals over which our function here, F of X is increasing and kind of gave up where we have that X is greater than or equal to zero. So I function here is F of X is equal to the integral, going from zero to X of one plus T over one plus t squared DT sort of find the intervals for which the function is increasing. We want to find the interval for which the derivative. So if we take the derivative D D X. Of our of our functions, so the derivative of fx, we want this to be greater than zero. So here we can just go ahead and substitute in the derivative and then make use of the first fundamental theorem of calculus because the derivative of and then a girl, it's just gonna be equal here to one plus acts which go ahead and put in X 14 and we get that are derivative here is just to be equal to one plus x over one plus X squared. Now, since the denominator, the denominator here is always going to be positive. Therefore the first derivative is always going to be positive whatever the numerator is positive. Therefore, since the range of the function is X greater than or equal to zero, we have that F of X is increasing for X greater than or equal to zero. Now to find the intervals for which our function is concave up. Want to find the interval over which the second derivative is positive. So we take the derivative of the derivative and when we want that greater than zero. So we already saw for the first derivative. So the second derivative here is again the derivative take the derivative. DDX of our first trip. That's going to be of one plus X over one plus X squared. And this is going to give us um eh well at one plus x ups a um a 1/1 plus X squared minus two X times the quantity one plus X all divided by the quantity one plus x squared quantity squared. And now the interval here for which our function is concave up is when are stuck in derivative is positive. So we have that the one plus 1/1 plus X squared is greater than two. X times one plus X over one plus X squared squared, which gives us that one plus X squared is going to be greater than two X plus two X squared which gives us that one minus two, X minus X squared is going to be greater than zero. Um We can then use the quadratic equation here to sell for X and we get that what we have, let's see, we got to get a negative negative two and then plus or minus the square root of negative two squared minus four times negative one times one. All divided by two times negative one, which gives us a positive two years. We have to plus or minus the square root of four uh plus four over um over negative two, which gives us negative one plus or minus route to. So since the function, the range of the function is X graded equal to zero, the function is going to be concave up. Um For the interval zero less than or equal to X, which is going to be less than while the square root of two minus one. Oops. What is what? All right. Take care. Been crying.

Question 44 says that if A is greater than zero and B is greater than zero show that f of X equals a times one minus E to the negative B X. Um is everywhere increasing and everywhere concave down. Um So testing if it's increasing, we take f prime of X, which is equal to one plus A B. E to the negative B. X. And this function, we know that A and B are greater than zero E to the negative Bx is always positive. Therefore F prime of X is always greater than zero, which means it's always increasing. Now taking f double prime of X. Furqan cavity, we'd have negative A B squared E to the negative B X. Again A and B are positive. Each of the negative B x is positive. So this negative out front means we're always negative and a negative fto prov X means we are concave down And those are your answers for question 44.

So we have a of X is equal to that integral up top there, and they want us to match up these properties. So for this first one, it's saying a is decreasing. Well, if a is decreasing, um, well, let's think about. So if we were to have that, that means a prime of X is going to be less than zero. Well, if a prime of X is less than zero, well, then if we take the derivative of this using the fundamental therm of calculus that is going to be equal to a prime of X. So actually, maybe I should write this in the corner, appear so a prime of X is just gonna be equal to app. So if this is less than zero, then that means f of X is us in zero. And if f of X is us in zero, then that means we are below the X access. So a and I go together. So let's go ahead and cross those off. Now we have that A has a local maximum. Well, for a to have a local maximum, that means that if we had so let's say f prime of C is equal to zero. Then we want f prime to the left of this. So at some point A, we want this to be, uh, increasing into it. So it wants to be positive increasing into it. And then we want f prime of B two b Decreased air bones are not EPS. Let me get rid of those because we're talking about a still so it would be the same idea. Just have the A is there instead. So if so, I'll just say a local max. So that means a prime of X to the left of this point is going to be increasing because we're increasing into the point. And then we're decreasing after so a prime of X is less than zero. Well, kind of like what we said before. We already know a prime of X is equal to f. So then that means over here F of X is greater than zero. And then on the other side, F of X is less than zero. So this means we are above the ex access and it's positive. And then over here were below B X access, and that's negative. So it looks like B and two are going to be a pair also. So maybe I should just make this a little bit bigger. I'll put all the solutions in this box right here. So those a repair and then we have be and to our repair. All right, so let's go ahead and cross those two off now A is con cave up. Well, if a is con cape up, that means so I'll come do this down here. This means a double prime of X is going to be greater than zero. So we already know that a prime of X is equal to ffx. So then that means a double prime of X is going to be equal to f prime of X. So if this is greater than zero, then that means f prime of X is very than zero. So we have zero here, and then this implies f is increasing. So that's four. So we can do C and four together. So let's go ahead and write that up. Here s O. C. And four. And then lastly Well, just by process of elimination, this means that okay, D and three will go together, But we can also go ahead and actually explain why that ISS Let's go ahead and do that down here. So it says a goes from concave up to con cave down at some point. So if that's the case con cave up, so that means a double prime effects is going to be greater than zero. Like we said from part C, And then it's going to go to Con Cave down, which means is less than zero. But again, we know a double prime of X is equal to f prime. So that means here f prime of X is greater than zero. And then this goes to F prime of X being less than zero. And if we are increasing into a point and then decreasing after that is a local Max. So that is why we can match, uh D and three together, but yeah, so these are going to be our four pairs or this one


Similar Solved Questions

5 answers
What volume of material is removed from solid sphere ofradius Zr by drilling hole of radius through the center? Verify that the shell" and "disc" method give the same resull:
What volume of material is removed from solid sphere ofradius Zr by drilling hole of radius through the center? Verify that the shell" and "disc" method give the same resull:...
5 answers
Of the Consider Cr##t -ArLMAM C-Hsle) Chooseoneor the U U 1 five balanced 93 I aillmc I = L Treactions listed bclow; <12024 ese reaclant Normallv; would be the limiting reactant Ssjjxjue 5i H Sce Periodic Table ofany moles rcactions See Hint
of the Consider Cr##t -ArLMAM C-Hsle) Chooseoneor the U U 1 five balanced 93 I aillmc I = L Treactions listed bclow; <12024 ese reaclant Normallv; would be the limiting reactant Ssjjxjue 5i H Sce Periodic Table ofany moles rcactions See Hint...
2 answers
Teaeoa aatl 7at teetat 009tdtia47uu tatu Murr Teullinm Us 120la" #en 4eLa lethe 6en0n44da ht eltia whjt 17{60n turn40
Teaeoa aatl 7at teetat 009tdtia47uu tatu Murr Teullinm Us 120la" #en 4eLa lethe 6en0n44da ht eltia whjt 17 {60n turn40...
5 answers
10 of 12 (7 complete)uted, with mean u = 90 and standard deviation & = 16. Compute the probability POX < 110)
10 of 12 (7 complete) uted, with mean u = 90 and standard deviation & = 16. Compute the probability POX < 110)...
5 answers
66. Define tn inductively by t 1 and tn+l tn/(1 + t8), where B is fixed, 0 < 8 < 1. Prove that Lk-I tn converges: [Hint: Find a constant C such that tn < Cn-1/8 ]
66. Define tn inductively by t 1 and tn+l tn/(1 + t8), where B is fixed, 0 < 8 < 1. Prove that Lk-I tn converges: [Hint: Find a constant C such that tn < Cn-1/8 ]...
5 answers
Question 41 ptsThe ethanol that is added to gasoline as a "renewable fuel" component is produced by fermenting corn glucose (equation is below): Determine AG? at 421 KCbH12Oslaq)2 CzHsOH()2 COzlg) AH?82.4k; 4S8460 J/K358 kJ276kJ+358 kJ+81.9kJ194kJ
Question 4 1 pts The ethanol that is added to gasoline as a "renewable fuel" component is produced by fermenting corn glucose (equation is below): Determine AG? at 421 K CbH12Oslaq) 2 CzHsOH() 2 COzlg) AH? 82.4k; 4S8 460 J/K 358 kJ 276kJ +358 kJ +81.9kJ 194kJ...
5 answers
Identify the region of this molecule that is hydrophobicchz Nc ctOFPCHz CH CHzCe0 C0Typexltetelia search
Identify the region of this molecule that is hydrophobic chz Nc ct OFP CHz CH CHz Ce0 C0 Typexltetelia search...
5 answers
Find an squatiun Uf the clrcle whose diameter has endpoints (-2, ~5) &d (-6,3).8 0D-D
Find an squatiun Uf the clrcle whose diameter has endpoints (-2, ~5) &d (-6,3). 8 0D-D...
5 answers
Provided the initial amounts of reactant A and product B (box on the left), complete the boxes on the right for a given magnitude of the equilibrium constant; Kc: (aq) = B (aq) Kc <1 Kc = 1 Kc >>1
Provided the initial amounts of reactant A and product B (box on the left), complete the boxes on the right for a given magnitude of the equilibrium constant; Kc: (aq) = B (aq) Kc <1 Kc = 1 Kc >>1...
5 answers
Suppose thatcenain company; the relatlonshlp between the price per unit _ of Its producr and the weekly les voiumetnousands dollars,given byp +5Solve thls differentlal equation Y = 16 when527 .Need Help?Ue lte
Suppose that cenain company; the relatlonshlp between the price per unit _ of Its producr and the weekly les voiume tnousands dollars, given by p +5 Solve thls differentlal equation Y = 16 when 527 . Need Help? Ue lte...
5 answers
What alkane is needed to make each alkyl halide by radical halogenation?
What alkane is needed to make each alkyl halide by radical halogenation?...
1 answers
Using the table of conversion factors on the inside front cover, convert (a) 760 miles/h to m/s. (b) $921 \mathrm{kg} / \mathrm{m}^{3}$ to $\mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}$ (c) $5.37 \times 10^{3} \mathrm{kJ} / \mathrm{min}$ to hp.
Using the table of conversion factors on the inside front cover, convert (a) 760 miles/h to m/s. (b) $921 \mathrm{kg} / \mathrm{m}^{3}$ to $\mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}$ (c) $5.37 \times 10^{3} \mathrm{kJ} / \mathrm{min}$ to hp....
5 answers
Simplify the trigonometric expression: seck()_1 sec2(x)
Simplify the trigonometric expression: seck()_1 sec2(x)...
4 answers
Solve the equation 3964` 40457 for the numerical value of a. Discuss either the relevance or the lack of relevance of your answer to pait as a counterexample to Fermat' s Last Theorem (which says that a" + bn = c" has 10 positive integer solutions for &, b, and n if n23).
Solve the equation 3964` 40457 for the numerical value of a. Discuss either the relevance or the lack of relevance of your answer to pait as a counterexample to Fermat' s Last Theorem (which says that a" + bn = c" has 10 positive integer solutions for &, b, and n if n23)....
5 answers
(b)T/2 % sin(x) cos() dx
(b) T/2 % sin(x) cos() dx...
5 answers
7_ Given: 2 =6 Cos 3 +isin 3 w =2 cos T + isin 6 6(5 points) A Find zw. Leave the answer in the form a + bi,(5 points) B. FindLeave the answer in the form a + bi.
7_ Given: 2 =6 Cos 3 +isin 3 w =2 cos T + isin 6 6 (5 points) A Find zw. Leave the answer in the form a + bi, (5 points) B. Find Leave the answer in the form a + bi....

-- 0.049953--