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The function f(c) 2c8 271? + 841 5 has two critical numbers:The smaller one is xand the larger one isCalculatorSubmit Question3E0"888t 362WNab...

Question

The function f(c) 2c8 271? + 841 5 has two critical numbers:The smaller one is xand the larger one isCalculatorSubmit Question3E0"888t 362WNab

The function f(c) 2c8 271? + 841 5 has two critical numbers: The smaller one is x and the larger one is Calculator Submit Question 3 E0 "888 t 3 6 2 WN ab



Answers

In Exercises $29-58,$ find the min and max of the function on the given interval by comparing values at the critical points and endpoints.
$$
y=3 e^{x}-e^{2 x},\left[-\frac{1}{2}, 1\right]
$$

Talk about Question 57. We need to find a maximum and minimum value of the function and the given interval, using the behavior at critical and and points. So let's start first by finding the critical points by differentiating and equating to zero three comes outside. It is a constant the physician off it is to the politics as it is to the politics. Only minus the transition of it is to the power to exist. It is to the power to extremes different station off power with the different station off to exist just to we quit this derivative to zero. So we have three times areas to the products. When it's two times areas to the part two X is equal to zero To simplify, the things will substitute areas to the bar xsd And if we square both sides, we have it is to the part two weeks as the square. This means that we have three t minus two t square as zero, so that will help us factories. Er, you have three times three minus duty is zero. It means that he is zero and they would be three over these other two values off T, but we're not interested in tea. We have to find X. So it is to the poor. X is either zero or it is to the poor exes. Either three or two. This is just not possible because it is to the power X is always positive. Uh, this value gives us taking natural law. Both sides, we have the value of excess. Ln three over. Do so. This is one value. And, uh, let's see. Let's find the sign. Change about l entry or two off the derivative t by. Over. T x. This is the derivative over here. Mhm on. If he put the value off T less than three over to that will be positive on for liberated than to you over towards three. Over towards negative. So it means that we have a local Maxima at Ellen. 3/2 on. If you want to find the value off, why, then we substitute us. We have three times areas to the power. Ellen. Three or two minus. Here is to the poor. Two times. Ellen, three over to. So this value comes somewhere around 2.25 This is the approximate value of why our tax equal to Ellen Tree over 12 Elementary over two is somewhere on find 405 Now let's talk about it's 10 points as well. So when points are minus one or two, this is 19 point. So the while the value off why is three times it is to the ball minus one or two, minus areas to the ball minus 2/2 is nothing but minus one. So negative one or two value comes somewhere around 1.45 on an X equal toe. The other end point is one but X equal to one. We get y as three ee days to the par minus on minus serious to the power it will be easy is to the fall of one year start minus one. It's easy is to the fall of one minus Edie's to the power to on this comes somewhere around zero point 766 Really, the system minimum value and among these two among 1.45 and 2.25 days is the maximum value. So to sum it off, we have a maximum value of critical point. So we have a Maxima off 2.25 at X. Somewhere at X value is approximately 0.405 and we have a minimum. We have a minimum value or the MINIMA off 4.766 are one of the endpoint, which is X equal toe one. So this is the final answer.

Looking at 37 you find are credible values, but you're going to be wide. Crime is equal to Isay to the fore minus eight, setting that equal to zero of solving Racine. Yes, Z to the fourth is equal to 16 which clearly Z is equal to bust or minus two. Well, both of those air in our rain. So we're going to test all of these values. 1,000,000. Why negative three. We get a value of negative threes clinging. Why at three. Forget about your positive three, but flooding in. Why negative too? Get negative 1 20 years. And why at two was positive. 1 28 looking that maximums are clearly going to be 1 28 9 minutes from the negative one.

Looking at Chapter five, Section one Problem number 14 were given a function f of X This 2.9 plus by puts xx my this 1.3 x squared and we arrested the three things were asked to find the critical number and then find where F of X is increasing intervals where it's increasing and intervals of decreasing. So we start by finding the derivative. So switch pen colors here and say that that prime of X this sequel to so the derivative of a constant, of course, zero less, uh, 5.6 then using the power rule. So we're gonna multiply by two so 2.6 and lose one on exponents. And so the first thing we want to do to find our critical number is find the derivative and said it equal to zero. So what? I said by 0.6 minus 2.6 x equal to zero, I'm gonna get X equal to, So that'll be 5.6 over 2.6, which reduces to 28 over 13. So we say that 28 over 13 is a critical number. Could abbreviate that. So we'll say our critical member and then with our critical number, what we can do is set up a signed shirt, so make a little straight line here and backed my pen. So are signed. Chart is using the critical number 28 over 13. And we got that number by setting a prime equals zero star sign charges related to F prime Then So then we're gonna look at a value to the left of 28 over 13 and plug into a prime and see if it's positive or negative. So if I take zero, for example, easy one less than 28 over 13 I'm gonna get popped. Positive. 5.6. So if I'm positive, that tells me that I must be increasing over that interval. Evan. Likewise, if I try a number to the right of 28 over 13 so I could pick 10 let's say, um, than 5.6 minus 2.6 times 10. It's gonna be a negative. So if we're negative, then we say that the function f of X must be decreasing over that interval. So writing out our solutions silver part, eh? We completed that here that we got that 28 over 13. It's our critical number for part B. They're asking us the last list intervals where F of X would be increasing. So we can say effort. Becks. Yes. Increasing over the interval from negative infinity to positive. 28 over 13. And we can say part seeing then that f of X is decreasing over the interval, starting at 28 over 13 to infinity, and we have it.


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