One of the important equipment used to confirm mathematical results in Calculus is a Mean Benefit Theorem of which states that if f(x) is described and is steady on the period of time [a, b] and is differentiable on (a, b), there is also a number c in the time period (a, b) [which means an important b] such that,

f'(c)=[f(b) -- f(a)]/(b-a).

Example: Consider a function f(x)=(x-4)^2 + you on an period of time [3, 6]

Choice: f(x)=(x-4)^2 plus 1, given interval [a, b]=[3, 6]

f(a)=f(3)=(3-4)^2 + 1= 1+1 =2

f(b)=f(6)=(6-4)^2 plus 1 = 4+1 =5

Using the Mean Value Theory, let us get the kind at some point vitamins.

f'(c)= [f(b)-f(a)]/(b-a)

=[5-2]/(6-3)

=3/3

=1

Therefore , the derivative at city is 1 . Let https://iteducationcourse.com/remainder-theorem/ find the coordinates of c by simply plugging on c inside the derivative on the original equation given make it equal to the result of the Mean Worth. That gives you,

f(x) = (x-4)^2 +1

f(c) sama dengan (c-4)^2+1

= c^2-8c+16 plus1

=c^2-8c+17

f'(c)=2c-8=1 [f'(c)=1]

we get, c= 9/2 which is the x value of c. Plug in this benefit in the original equation

f(9/2) = [9/2 supports 4]^2+1= 1/4 +1 = 5/4

so , the coordinates from c (c, f(c)) is certainly (9/2, 5/4)

Mean Value Theorem intended for Derivatives claims that whenever f(x) is known as a continuous labor on [a, b] and differentiable in (a, b) then there is also a number vitamins between a and udemærket such that,

f'(c)= [f(b)-f(a)]/(b-a)

Mean Value Theorem for Integrals

It suggests that in the event f(x) may be a continuous party on [a, b], then there is also a number c in [a, b] in a way that,

f(c)= 1/(b-a) [Integral (a to b)f(x) dx]

This is the First Mean Value Theorem pertaining to Integrals

From your theorem we can easily say that the average value in f upon [a, b] is achieved on [a, b].

Example: Permit f(x) sama dengan 5x^4+2. Identify c, so that f(c) is definitely the average value of farrenheit on the span [-1, 2]

Answer: Using the Mean Value Theorem for the Integrals,

f(c) = 1/(b-a)[integral(a to b) f(x) dx]

The regular value of f within the interval [-1, 2] has by,

sama dengan 1/[2-(-1)] major (-1 to 2) [5x^4+2]dx

= one-half [x^5 +2x](-1 to 2)

= 1/4 [ 2^5+ 2(2) - (-1)^5+2(-1) ]

= 1/3 [32+4+1+2]

= 39/3 sama dengan 13

As being f(c)= 5c^4+2, we get 5c^4+2 = 13, so vitamins =+/-(11/5)^(1/4)

We have, c= 4th root of (11/5)

Second Mean Value Theorem for the integrals state governments that, In the event f(x) is continuous when using interval [a, b] after that,

d/dx Integral(a to b) f(t) dt = f(x)

Example: get d/dx Important (5 to x^2) sqrt(1+t^2)dt

Solution: Making an application the second Mean Value Theorem for Integrals,

let u= x^2 that gives us y= integral (5 to u) sqrt(1+t^2)dt

We all know, dy/dx = dy/du. dere. dx sama dengan [sqrt(1+u^2)] (2x) = two times[sqrt(1+x^4)]

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