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Dctermide the >-valuos ot mlative @xUema and ncinis Inteetroe the function f() hnlerVa (0.10). wlch thc Graph of f () Prowidod Fuvide cxnlanatlon (0{ Your an wwc...

Question

Dctermide the >-valuos ot mlative @xUema and ncinis Inteetroe the function f() hnlerVa (0.10). wlch thc Graph of f () Prowidod Fuvide cxnlanatlon (0{ Your an wwcre Anqwor; Mithout axclanatlon olntsLocalvlnImzF (~) Teles 8~ 6co" ^egatie Pottty Doz Lz0 2e Lall= 'th~ J ELh Maxima L4 {) Vcllce S Gro Postt_ Fo ~ ~Nl_ 450-28 #lc values; hinll ~ 472 IcolPointls) - Inflection; 414 Follou? 4 ~aue: F( ) Ssitcke 2~ Lecn #~osc Arc Fointa 0t Mf(ectiov_bl There Is a function g(x) that different

Dctermide the >-valuos ot mlative @xUema and ncinis Inteetroe the function f() hnlerVa (0.10). wlch thc Graph of f () Prowidod Fuvide cxnlanatlon (0{ Your an wwcre Anqwor; Mithout axclanatlon olnts LocalvlnImz F (~) Teles 8~ 6co" ^egatie Pottty Doz Lz0 2e Lall= 'th~ J ELh Maxima L4 {) Vcllce S Gro Postt_ Fo ~ ~Nl_ 450-28 #lc values; hinll ~ 472 Icol Pointls) - Inflection; 414 Follou? 4 ~aue: F( ) Ssitcke 2~ Lecn #~osc Arc Fointa 0t Mf(ectiov_ bl There Is a function g(x) that differentiable on the open Interval (0,10) that has derivative g' (x) defined by the equation g' (x) (r - 6)(* - 9) " f"(x) Determine the X-values of all relative extrema of g(x) on (0,10). Provlde justification for vour answers. Answers wlthout Justification will receive no polnts_



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Think About It In Exercises 79 and 80 , the function $f$ is differentiable on the indicated interval. The table shows $f^{\prime}(x)$ for selected values of $x$ . (a) Sketch the graph of $f,(\text { b) approximate }$ the critical numbers, and (c) identify the relative extrema.
$$
f \text { is differentiable on }[-1,1]
$$
$$
\begin{array}{|c|c|c|c|c|}\hline x & {-1} & {-0.75} & {-0.50} & {-0.25} \\ \hline f^{\prime(x)} & {-10} & {-3.2} & {-0.5} & {0.8} \\ \hline\end{array}
$$
$$
\begin{array}{|c|c|c|c|c|c|}\hline x & {0} & {0.25} & {0.50} & {0.75} & {1} \\ \hline f^{\prime(x)} & {5.6} & {3.6} & {-0.2} & {-6.7} & {-20.1} \\ \hline\end{array}
$$

So we're gonna use this data to help sketch our graph. So what this is gonna look like is um X equals zero. So we're going to get f prime of X is 3.14. So zero. We're gonna get a slip of about 123. So it's going to be about something like this perhaps. And then um at pi over six which is about one a half, we're going to get a negative slope negative Um .23. So it's gonna be very close to zero but it's going to be slightly negative. So at about 1/2 you see it's already going to be slightly negative And then at power four Which is going to be slightly less than one. About 3/4 it's gonna be very steep. So we see that it's going to be looking like something like this And then we see it reaches the lowest negative about Pi over three. That's about one, that's gonna be that low. And then we see it's going to come back so it's going to be this oscillating function. So that's our final answer.

That's my sister Christa grabs looked like a on track automatic function. And when he started from flattered from in from the grab, we can see the critical numbers are approximately 0.5, 1.5 and 2.5. And from welcome. You can see that with any relative minimum AC 3.55 2.5 I I had 25 points at 2.5 and yeah, 1.5 negative. I the relative makes him, um, in the relative minimum.

First we want to find all extreme points. So since from 0 to 2 we see that we have a positive slope And then from at to the slope is zero And then from 2-3 we have a negative slope. That means that it's going to be looking like this. So then we have a local maximum at two. Then it goes to three, it's negative, it continues to three, it's negative, it goes from 3-4 being negative And then at four it reaches zero. But then from 4 to 6 it goes negative. So we see that there is no maximum there it's just going to be a maximum at X equals two. But then in terms of the con cavity we see that we are concave down because it's negative and it's negative the whole time until we reach three at three, after three it goes positive. So that means that X equals three is going to be our inflection point. And then when we reach four It's at zero again and then it goes negative. So four is also actually going to be an inflection point, because then it goes concave down again, so three and four are inflection point.

Hello. So here we are considering the function G of X is equal to the integral going from zero acts of F F T D. T. So here we have the fft oscillates about the line Y is equal to two and the the X region going from zero 10. Now, since here the G Fx um with the integral going from zero X F F T D. T. Um That is just equal to F. Of X. So then G fx will have a local max where F goes from being positive to being negative. This occurs where X is going to be approximately equal to three five, seven and nine. Um And G will have a local minimum where F goes from being negative to positive. This occurs again, this here is um where actually is a local max And then X has a local men. Where um again f goes from being negative to positive. This occurs where X is approximately equal to 3.9. 5.9 7.9 And 9.9. Okay, so here she is going to have a local men. Okay, That's part A And then for part B. Well, the function G appears to have its global max at X is equal to nine. Um The point where the area above the curve minus the area below the curve is at its greatest and the function G appears to have its absolute men at X is equal to zero. So again, this is a global max and then the eps then the absolute men um access to go to zero since the area under the X axis is never greater than the area above it. So therefore the graph never gets back to its starting point. Okay. And then four parts see um well, the function G is going to be concave down anywhere where the graph of F is decreasing, so G is con caved down on the intervals from Um 0.5- 1.5. Um 2.5- 3.5. 4.5- 5.5. Um 6.5- 7.5. And from 8.5 To 9.5. And then um a rough sketch for d. We can make a rough sketch, a very rough sketch here. So a rough sketch of this graph is going to look something like this. So we here we have our y and I guess our T axes here. Um So we can have the T axis going from 1 to 10, we have 12 And so on. Up to 10. Um And the graph is going to look, something is happening here from the origin. Okay, so we come like this and then we start having these loops that look something, something like that. Right?


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