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Oint) Use the Second Derivative Test for local extrema to find any local maximum and minimum values for the function_f(r)=l-9x-6r" _XYou will receive partial c...

Question

Oint) Use the Second Derivative Test for local extrema to find any local maximum and minimum values for the function_f(r)=l-9x-6r" _XYou will receive partial credit if You instead using the First Derivative Test for local extrema and sign chart to determine each local extrema is either local maximum Of minimum_

oint) Use the Second Derivative Test for local extrema to find any local maximum and minimum values for the function_ f(r)=l-9x-6r" _X You will receive partial credit if You instead using the First Derivative Test for local extrema and sign chart to determine each local extrema is either local maximum Of minimum_



Answers

Use the Second Derivative Test to find the local extrema for the function.
$$y=x e^{x}$$

Mhm. Here we have F equals X times E. To the X. Taking the first derivative. Using the product rule we get one plus X. Either the X second. We have to be using the product will again we get two plus X. Neither the X. This the first derivative is zero when X is minus one, this is never zero. So that's the only time we have, we have a horizontal tangent. And if we plot this, we can see it looks like this. Now we take um take minus one and plug it into here. We see that we get you get that. This is positive. Mm hmm. Mhm Because obviously this is positive and this is positive also. So that means we have the second river, there is positive here, which means we have upwards curvature, which means that this thing is a minimum at -1. And in fact it is well no, it's not going to be a global minimum. Um because as X goes to minus infinity, what is the limit of that? I'm not sure, You see, I'm not sure what the limit of that is. As X goes to minus infinity um I think yeah, goes to zero. So this would be a a global max global minimum because this ass and talks to zero out here. Um because this goes to zero and this goes to infinity. So we need to take a limit. It turns out that the limit as x goes to infinity point of being zero

It's probably we have f of X equals X times either the minus X. Um which looks a lot like when you plot. It looks a lot like the previous problem. Um with X and Y switch with both Excellent and wise being switched. Um So let's see here taking first derivative we get the first derivative is x minus one times minus x. Second derivative gives us x minus to the minus X. So first we have to be zero and that's the only time when the zero is when X equals one because this is never equal to zero plugging in X equal one into here. You see that this thing is negative. So that means in the region where um the um the first revert of is zero. We have downwards curvature which means that um X1 is a local maximum. Then out here we have upwards curvature here we have an increasing function and here we have a decreasing function. Um And so here's our local maximum. This awesome talks to zero. So this is actually a global maximum. Or also

It's probably we have f of X equals X times either the minus X. Um which looks a lot like when you plot. It looks a lot like the previous problem. Um with X and Y switch with both Excellent and wise being switched. Um So let's see here taking first derivative we get the first derivative is x minus one times minus x. Second derivative gives us x minus to the minus X. So first we have to be zero and that's the only time when the zero is when X equals one because this is never equal to zero plugging in X equal one into here. You see that this thing is negative. So that means in the region where um the um the first revert of is zero. We have downwards curvature which means that um X1 is a local maximum. Then out here we have upwards curvature here we have an increasing function and here we have a decreasing function. Um And so here's our local maximum. This awesome talks to zero. So this is actually a global maximum. Or also

We need to find all relative extreme. I then use the second derivative test to see whether the it is a relative minimum or a relative maximum. So we have the function F of X is equal to X squared minus X. So to get the critical points, we find the first derivative and set it equal to zero. So the first derivative is two X minus one. So here we get that X is equal to one half. So this is our critical value and now are critical. Point is, so we have f at one half is equal. Thio one quarter minus one half, Just negative, one quarter. So we have the 0.1 half negative one quarter. Now we use the second derivative test to see whether it is a relative minimum or a relative maximum. So the second derivative is equal to two. And since the second derivative is greater than zero, you have a minimum


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