{Use of Tech} Critical points and extreme values

Question

a. Find the critical points of the following functions on the given interval. Use a root finder, if necessary.

b. Use a graphing utility to determine whether the critical points correspond to local maxima, local minima, or neither.

c. Find the absolute maximum and minimum values on the given interval, if they exist

h(x) (5-x)/(x² + 2x - 3) on [-10,10]

Accepted Answer

Welcome back, everyone. In this problem, we want to find the critical points, the local maximum value, and the local minimum value of the function H X equals 7 minus X divided by X + 3X minus 10 on the interval open bracket -10, 10 close bracket. Also, check for the absolute maximum and the absolute minimum if they exist. Now, there are a few things we need to establish about H of X, OK? Now, first, notice here that the denominator X2 + 3X minus 10 could be rewritten as X + 5 multiplied by X minus 2. And this is, this is pretty important because no, this tells us that H of X is going to be undefined at the point where our denominator is 0, which is when X equals -5. And X equals 2, OK. In other words, they are going to be vertical asymptotes at these points. Now that we know where it is undefined, OK, then we can go ahead and see if we can solve for our critical points within this given interval. OK. To find the critical points, we differentiate H of X and set it to 0, OK? So, at the critical points. The derivative of H of X will be equal to 0. Now, what's the derivative of HX? Well, we can differentiate this by using the quotient rule of differentiation, and when we do that, the derivative of H X will be equal to X2 minus 14 X minus 11 divided by X2 + 3 X minus 102. So we are going to set this to 0, and if this equals 0, then that means our numerator X2 minus 14 X minus 11 will be equal to 0 if we multiply through by our denominator. So now we can solve this by using the quadratic formula, OK? I know for the quadratic formula, it will be 1, B is -14, and C is -11. So that means then X is going to be equal to negative B, so that's -14 or 14, plus R minus B2-14 square minus 4 multiplied by A1 multiplied by C-11. And this is all divided by 2 multiplied by A, which is 1. Now, when we simplify here, we get this to be 14 plus or minus the square root of 196 + 44 or 240 divided by 2. And if we divide through by 2 and rewrite our root, then we should get X to be equal to 7 plus or minus 2 route 15, OK? Now, if we think about it here, only X equals 7 minus 2 route 15 lies in our interval, uh, -10 to 10 inclusive, OK? Thus, X must be equal to 7 minus 25, sorry, not route 5, Route 15. So this is our critical point. Now that we have the critical point, we've answered one part of our problem. Next, we need to figure out if, or we need to find a local maximum and local minimum values, OK. And then check to see if there are any absolute maximum or minimum on our intervals. Now here, OK, to figure that out. We can set, we, sorry, we can substitute the values for our endpoints and our critical point in the H, OK? So here. Evaluating H of X. A Critical points. And end points, OK. Then we're going to be finding H of -10. OK. Uh, age of 7 minus 2 route 15, OK, and age of 10. Now to find H of -10, we'll just substitute -10 into H of X for X, and when we do that, we should get 17 divided by 60. When X is 7 minus 2 route 15, H would be -4 route 15 plus 17 divided by 49. And when X equals 10 H would be -140. Now if we think about it here, when we ask ourselves about the behavior of our function, then as X approaches -5 or X approaches 2, OK, that is of vertical asymptotes, then H of X is going to approach negative or positive infinity. Thus, that means no absolute minimum or maximum exists. So that's an important point to note here. No absolute maximum or minimum exists, OK? So thus now we need to figure out which one of these. Uh, in other words, which one are either a critical points or end points are going to be a local maximum or minimum if they exist? Now here checking for a local maxima. For x less than 7 minus 2 root 15 or critical point, then our first derivative is going to be greater than 0. It's going to be positive. But on the other hand, when X is greater than 7 minus to root 15, then. Our first derivative is less than 0. The function is decreasing. Thus, since that's the case, that means X equals 7 minus 2, 15 is a local maximum, yeah, and we do not have a local minimum. Thus, to sum up all of what we've said here. OK. Because this is a local maximum. Then the maximum, well, I should probably write that in a better way here, but what we're saying that this point corresponds. 2 A local maximum which means that The value we got before for age of 7 minus 215. OK. -4 route 15 + 17 divided by 49 is the local maximum. Yeah, the value that it corresponds to is the local maximum. So now to summarize here. -4 route 15 plus 17 divided by 49 is the local maximum. Our critical point is +72 minus 2 route 15 and 4 route 15 + 17 divided by 49, and there is no local minimum, neither does an absolute maximum or minimum exist on the interval. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Critical Points

Local Extrema

Absolute Extrema

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