Problem

If $a$ and $b$ are positive numbers, find the max…

01:13

Problem 62 Hard Difficulty

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

$ f(x) = x - 2 \tan^{-1} x $, $ [0, 4] $

Answer

Absolute minimum value $1-\dfrac{\pi}{2} \approx -0.5707963268$ which occurs at $x=1$ ;
Absolute maximum value $4-2\tan^{-1} \left( 4 \right)$ which occurs at $x=4$

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Video Transcript

let's find the absolute minimum, an absolute maximum values of the function F of x equals x minus two, inverse of the tangent function on x defined on the close interval 04. So first we note that these functions continues over the interval 04. And that because the function universe of the tangent is defined on the real numbers and continue on the real numbers. So it's particularly continue defining continues on the interval 04. Remember that if we uh constrain the tangent function to the domain negative by half by half open interval both sides, then we have in verse well defined. Which is the function that we call tangent to the negative one of X, which is the inverse of the tangent function, which is then like this. So, as we can see, this function is defined and continues under real numbers. Well, so let's say here this is attendant to a negative one of eggs. So f being continuous and defined on a close interval, it attained its extreme values on that interval. And we know that those extreme values can be attained either on the end points of the interval or at critical numbers of the function F. So we get to start by calculating the first derivative of F, Which is 1 -2 times the derivative of the inverse of the tangent function of X is one over one plus six square. So this is equal to 1 -2 over one plus x square. And that's equal to one plus x square as a common denominator. In the numerator we have one plus x square minus two. So the first derivative of f is equal to x square minus one over x square plus one. And uh the denominator we have in this expression is never cereal. In fact it's always positive X square plus one. And so we have that these derivative exists for every uh X in the close interval 04. For that reason we can say that The only critical numbers of f in the interval 04, our total this effects in that interval for which the first derivative is equal to zero. And so we get to solve this equation here and we start with that equation. And because we know this first derivative is equal to the expression here is equivalent to saying that x squared minus one over x square plus one is equal to zero. And this is the same as s square plus one. Sorry, X squared minus one equals zero. That is a square equal one. And from here we know that it's can be negative one, or x can be one, but in this case The Value Native one is not In the interval 04. Then we can say that the only critical number of the function F In the interval 04 is X Equal one. So we have this Google number and the end points of the intervals as candidates for the function attained its extreme values. So we can say that F at the left zero is 0 -2, Inverse of the tangent function at zero. And because the tangent of 00, we get this is equal to zero. And at the writing .4 the function is able to form -2. Universal detergent at four. And is a calculator. We found that these values are approximately equal to one 34 83, 64, 6, 7, 2, 7. And finally F at the critical number x equal one is one minus two. Universe of the tangent at one. And because we know the changing of 5/4 is one. This give us one minus two times by fourth Which is one by half. And that's approximately equal To negative 0.57 0796. Uh 3- 6, 8. Okay, so these are the candidates for string values and functions of the largest of the three values. Is this one here. So it must be The absolute maximum value of the function over the interval 04. And it is occurring at the writing .4. So I will write this here. Absolutely maximum value of the function attained at the right important football. Now, the smallest of the three values is this one here one minus five half. And that attained at the critical number one. So this must be the absolute a minimum value of the function over the interest rate of four. So we write the answer following this observation here. So the the absolute minimum value of the function f uh in sorry on On dangerous 04 is this one here is one minus by health Which is a approximately equal to negative 0.57 0796 3268. And that absolute minimum of course at one. That is a critical number now. 40 absolute maximum Of the function on the interval. 04. He is this one here uh for minus two. Reverse of the tangent function at four which is approximately equal To 1.34836 for 6727. In that absolute maximum value value I forget here while you which of course at de critical number X equal one And I think I hear a road one which is four. It's clear. Look at this, it's just value here. This all here. Is that? Uh so sorry nice one. It was okay. And stopping a bit here. Uh critical number is one. We found it here above. Let's see it again. Critical number is one. So that that value one we have the absolute minimum. So the absolute minimum it's one minus pi half which I presumably equal to this value. And that of course are critical number one was perfect. Now for the absolute maximum value is four. This one here for my institute universal thing that four which presumably equal to that and that occurs. It's not a critical number repeating what it was love. It happens at four. That is the right and point of the interval. So here court at the right endpoint X equal for sorry for that. So this is a great answer and we recall a little bit what we have done here. So the first thing we got to do is to verify that the function is continuous on the clause interval that tells us that the extreme values of that functioning system, that interval are attained either at the end points of the interval or at critical numbers of the function. We calculated the first derivative of the function. We saw that that exists for every argument X in the interval 04. And for that reason the only critical numbers of the function of the in the intervals here for are those were the vax for which The first of the TV- zero we started that equation and there are two solutions native 11 but one of them namely X equal 91 is not in the internal steerforth. So the only critical number of the function f within the interval zero for is X equal one. So we have the critical number one. And the endpoints here. four. So we evaluated at those three points. And with that we found Like the absolute maximum value of the function over the Angels three or 4 which corresponds to the largest value calculated not just of the three values calculated uh Is 4 -2 tangent inverse at four, which is in fact the value that of course at Mexico four. Mhm. So the absolute maximum value Of the function over in two or 3 or four is 4 -2. Universal attended at four that which is approximately equal 1.34836467-7. And it occurs at the writing point of the interview that is at four, That's the largest value we found. And the smallest value, which is uh one minus by half, Which is approximately equal to negative 0.5707963268 Is then the absolute minimum value of the function over the interval 04 and it is attained or of course at the critical number X equal one. And so this is the final answer to the problem.

Oswaldo J.

Universidad Central de Venezuela

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Maximum and Minimum Values

Problem

If $a$ and $b$ are positive numbers, find the max…

Problem 62 Hard Difficulty

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

$ f(x) = x - 2 \tan^{-1} x $, $ [0, 4] $

Answer

Absolute minimum value $1-\dfrac{\pi}{2} \approx -0.5707963268$ which occurs at $x=1$ ;
Absolute maximum value $4-2\tan^{-1} \left( 4 \right)$ which occurs at $x=4$

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Video Transcript

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Absolute minimum value $1-\dfrac{\pi}{2} \approx -0.5707963268$ which occurs at $x=1$ ; Absolute maximum value $4-2\tan^{-1} \left( 4 \right)$ which occurs at $x=4$

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

Absolute minimum value $1-\dfrac{\pi}{2} \approx -0.5707963268$ which occurs at $x=1$ ;
Absolute maximum value $4-2\tan^{-1} \left( 4 \right)$ which occurs at $x=4$