#mean value theorem

[quantum-friend-theory]

Given a function f(x) which is continuous on an interval [a,b] & differentiable on its interior, the Mean Value Theorem guarantees there exists a value c in that interval such that the line tangent to the function when x = c is parallel to the line drawn between ( a , f(a) ) and ( b, f(b) ). (GIF Description: A function f(x) is plotted from x = a to b, and a red line is drawn b/w its endpoints at ( a , f(a) ) & ( b , f(b) ). A point tracing f(x) carries a compass that records the tangent's slope relative to the red line. When their slopes are equal, x is labeled c. In the first GIF, the function has a single maxima. In the second GIF, the function has multiple extrema.)

[studyswamp]

i feel like it’s that part of the year where im more motivated to go outside and enjoy the weather than i am motivated to study! oh well, we push on with coffee and benadryl...

[imanundertalelover-blog]

Formula Compilation - Calculus I <2>

Derivative rules and mean value theorem.

[ijtsrd]

A Note on the Generalization of the Mean Value Theorem

by Yeong-Jeu Sun "A Note on the Generalization of the Mean Value Theorem"

Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-5 | Issue-1 , December 2020,

URL: https://www.ijtsrd.com/papers/ijtsrd38258.pdf

Paper URL : https://www.ijtsrd.com/mathemetics/calculus/38258/a-note-on-the-generalization-of-the-mean-value-theorem/yeongjeu-sun

callforpapermedicalscience, medicalsciencejournal, manuscriptsubmission

In this paper, a new generalization of the mean value theorem is firstly established. Based on the Rolle’s theorem, a simple proof is provided to guarantee the correctness of such a generalization. Some corollaries are evidently obtained by the main result. It will be shown that the mean value theorem, the Cauchy’s mean value theorem, and the mean value theorem for integrals are the special cases of such a generalized form. We can simultaneously obtain the upper and lower bounds of certain integral formulas and verify inequalities by using the main theorems. Finally, two examples are offered to illustrate the feasibility and effectiveness of the obtained results. 

[mathplane]

[ahhhhhpplied-math]

STEP BY STEP guide to prove the Extreme Value Theorem

Definition of the Extreme Value Theorem: If the function f is continuous on [a,b], then there exists constants c and d in the interval [a,b] so that f( c ) is the absolute maximum of f and f(d) is the absolute minimum of f on [a,b].

Sounds *gross* eh? All of that just means given a continuous function f, we can pick an interval [a,b] and we can always find a local maximum and a local minimum in that interval. This is a common theorem covered in first-year calculus courses.

You might think that it’s obvious. 

But how do we prove that it’s true for all x? There are an infinite amount of x that we can sub into function f. It’s rather silly to test them all out. It will take us an infinite amount of time, and idk about you but I’m too lazy for that. So, we will prove this theorem, mathematically. 

We will prove this theorem by infinitely dividing the interval into halves. I have the written formal proof below, but it’s not important if you just want to know the basic idea of the proof. 

Step by step guide:

We will divide up our closed interval [a,b] into 2 parts.

We then look at the local maxima of each interval. 

We then pick the interval containing the maximum value of the original interval to be our new interval.

Wash, rinse, repeat.

Notice that everytime we split up our interval, the new one gets smaller. And when we repeat this process an infinite amount of time, our interval gets *REALLY, REALLY, REALLY* small. we want to divide it up to the point where the interval is reallyyyyyy closed to completely *disappear*. This is a very abstract thought. 

Remember that while we’re dividing up interval, we’re also choosing the one where the local maximum of the original interval is. So we’re approaching a value where f reaches its maximum. We won’t be arriving exactly where the maximum is ( because then the interval is non-existent ), but we’re getting arbitrarily closed to it. We can call this point c, and the maximum value f( c ). Because the function is continuous on the interval, that point we’re approaching exists and is the maximum. The existence of the minimum can also be proven this way.

Ok so what the fuck did i write in the proof?

The proof I wrote below is a bit more technical, but using the same idea. We divide up the interval n times. Let M be the least upper bound of f(x), meaning f( c ) is smaller or equal to M. Using the epsilon-delta definiton, I proved that f( c ) is larger or equal to M. But for these 2 statements to both be true, f( c ) must be M, the supremum of f(x) on the interval [a, b], and because f( c ) is on f(x), it’s the local maximum. 

coolio. 

[balajirao1980-blog]

VERIFY MEAN VALUE THEOREM

[cheappackersmoversz]

Intermediate Valorize Theorem for Derivatives

Introduction to minion value theorem for derivatives:Medium value theorem says that ' A continuous workings doing a blank and bounded interval attains every value between irreducible two given points among the range. ' Look at the following example,Let f: ]0, 5] `->` `RR` be defined by f(x) = `1\(x-3)` + 2 for x `!=` 3.We thimblerig f(0) = `5\3` and f(5) = `5\2`. Now we have `5\3` `Here f is not continuous. Along these lines cue of f is very important to garnish inteermediate value theorem.But once you learn ' intermediate value theorem for derivates' then you will be astonished to see that though we are not decided pertinent to continuity of a function, inner man satisfies the intermediate value figure. This happens so as to circa special functions. Those are differentiable functions and we will be message about the intermediate value property for their derivatives.<\p>

Statement of Intermediate Value Theorem as long as Derivatives:Intermediate market value theorem for derivatives is above known as ' Darboux Theorem '. This theorem roughly says that derivative in relation to a differentiable function satisfies intermediate value properrty on a level though it may not be continuous.Folderol: Let f: ]a, b] `->` `RR` be a differentiable function. If f^l(a) `Proof: Define g: ]a, b] `->` `RR` by means of g(mark) = f(x) - `lambda`x. Plus g is differentiable and g^l(a) `` 0. So kilohertz is decreasing at a and increasing at b. Also seeing as how g is continuous on ]a, b], it attains its extremum. But by primarily two conditions, the extremum is attained in the interior of (a, b). Repression it be attained at c. Then since c is an extremum point of g, we have g^l(c) = 0, that is f^l(c) = `lambda`.Thus we got a c `in` (a, b) such that f^l(c) = `lambda`. This completes the theorem.An Example for Intermediate Value Theorem in behalf of Derivatives:Define f: `RR``->` `RR` move f(x) = 2x sin(`1\x`) - cos(`1\x`) on behalf of x`!=` 0 and f(0) = 0. Does this function satisfy intermediate value property?Check this awesome icse syllabus for analyze 3 i recently used.By looking at the function we cant say directly whether it satisfies the ultimate property or not. But by some work we mass show that f is not continued at 0. So, can we conclude that ethical self doesnot satisfy middle stress property?. The answer is no. Observe that given f is a credited regarding g(terra incognita) = x2sin(`1\x`) for x`!=` 0 and g(0) = 0. So by means value theorem for derivatives, we can conclude that f satisfies intermediate value property.<\p>