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Find the absolute maximum and absolute minimum values of $f$ on the given interval.

$ f(x) = \ln (x^2 + x + 1) $, $ [-1, 1] $

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Absolute minimum value $\ln \left( \frac{3}{4} \right) \approx -0.2876820725$ which occurs at $x=-\frac{1}{2}$ ; Absolute maximum value $\ln \left( 3 \right) \approx 1.0986122887$ which occurs at $x=1$

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Wen Zheng

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Amrita Bhasin

04:36

Chris Trentman

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 1

Maximum and Minimum Values

Derivatives

Differentiation

Volume

Missouri State University

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Harvey Mudd College

Idaho State University

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Find the absolute maximum …

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let's find the absolute minimum absolute maximum values of the function ffx equal natural logarithms of X squared plus X plus one On the closed interval from negative 1 to 1. So first we know this function is well defined on that interval because first of all we see that X square plus six plus one can be written us eggs plus one half square which gives us X square plus X plus 1/4 Last 3 4th. Which in this case give us one And that's actually greater than or equal to 3/4. With this greater than zero. That is for all real number eggs. This expression is always positive. And so we can take the natural algorithm. So it will define and It is continuous on the interval -11. Because the composition of these two functions continued continuous in that interview. Mhm. So being continuous on a close interval means we know that the function attains its extreme values on that interval. And we know that uh those stream values are attained at either the end points of the interview or at the critical numbers of the function. So we calculate the critical numbers of dysfunction. And for that we need the 1st 1st derivative which is equal to the relative of the argument. Two X plus one over the argument. And that's the derivative. And because we said already that this denominator here is not zero. It's never serious. Always positive. This derivative exists for any eggs. And that means that Yeah. Or we can say that okay, the only critical numbers of F in negative 11 are those values of that interval for is the first derivative of F is equal to zero. So we got to solve these equation. First derivative of F equals zero. And that's equivalent to using the expression for the first derivative from here. To expose one over X square Plus eggs plus one include 0. And that's equivalent to two X plus one equals zero. And that's equivalent to x equal negative 1/2. So we can say that the only critical and of course this number is on the injuries in the interval. For me, I don't want to want the only critical number of F In danger. Well, 91 1 Is the value acts equal negative 1/2. So we know this is the only griddle number of the function. And then we add to evaluate the financial there. And at the end points 91 1. Okay, so we start with the endpoints. If a negative one is natural logarithms of uh 91 sq Plus 91 Plus one. These two values cancel out. And then we have natural algorithm of one in that zero. Then we have f. At one which is right on point. And that gives us natural algorithm of one square plus one plus one. That is natural over lower rhythm of three. And that is about not equal. Um That's about 1.2 or nine 861- two. 887. and finally f at the critical number of negative 1/2 is Natural algorithm of negative one have square plus nancy one half plus one. That's a natural or in the mouth three forth. And that's about -0.287 6820 73. Okay, so this is images of the importance and the only critical number. And for from these values we know that the largest value is Lanta algorithm of three which is 1.29. Um And that's the image at the writing point. That's then the absolute maximum value of the function on the interval -11. And it's attained to write a point of the interval. And the smallest of values here is negative 0.287 etc. There is not a natural algorithm of three falls. And that uh smallest value is the image of negative one. How which is the only critical number of the function. Then the absolute minimum value of the function over the interval negative 11 is Naturally with them of 3 4th. And of course at the critical number negative one half. So we can write the answer. Now then the absolute minimum value of F On the interval negative 1 1 is Uh natural algorithm of 3 4th which is about or Approximately equal to negative 0.287 68 20 73. And that absolute minimal value of course at the critical number. Yeah X equal negative 1/2. Then we have the answer for the absolute maximum the absolute maximum value of the function Over or on the interval from net in 1 to 1 is Natural algorithm of three which is about uh one Which is about one zero 1.09 86 1 to 2 887. And that absolute maximum value of course at the right point of the interval Which in this case is x equals one. So this is the final answer and a summary of what we have done here. We first verified that the function is continuous on the closed interval. Given in this case we also prove or verified that the functions well defined on that interval and it's also continues in that gloves into her. For that reason we know that the function attains its extreme values on that into and because we know the extreme values are obtained either the end points of the interval or at critical numbers of the function. We calculate the first derivative of the function and solve the equations first. There's a difficult zero. We verified that the first derivative exists for every number of value X. And into a negative 11. And for that reason the only critical numbers of the function in that interval are those values belonging to the interval for is the preservative zero. So we started stated the equation for preservative of X equals zero. And that's equivalent to the fact that X can be equal to negative one half. And from that we conclude that the only critical number of this function in the interval negative 11 is negative one half. So we got to evaluate the function at that critical number negative one half, and at the end points of the interval negative one and one we did that and the largest value found, which is then the absolute maximum value of the function over the interval nearly 11 was natural rhythm of three and that was obtained or of course at the writing point of the interval that is acceptable one and the smallest of the three of our values, which corresponds then to the absolute minimum value of the function over the interval Native 11 Is natural logarithms of 3, 4th and that Is attained at the critical number negative 1/2. And so this this is the final answer to this problem.

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