An Illustration of the Mean Value Theorem

One of the most important tools used to prove mathematical results in Calculus is the Mean Value Theorem which states that if f(x) is defined and is continuous on the interval [a,b] and is differentiable on (a,b), there exists a number c in the interval (a,b) [which means a b] such that, f'(c)=[f(b) - f(a)]/(b-a).

Example: Consider a function f(x)=(x-4)^2 + 1 on an interval [3,6] Solution: f(x)=(x-4)^2 + 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 + 1 = 4+1 =5 Using the Mean Value Theory, let us find the derivative at some point c.
f'(c)= [f(b)-f(a)]/(b-a) =[5-2]/(6-3) =3/3 =1 So, the derivative at c is 1.
Let us now find the coordinates of c by plugging in c in the derivative of the original equation given and set it equal to the result of the Mean Value.

That gives us, f(x) = (x-4)^2 +1 f(c) = (c-4)^2+1 = c^2-8c+16 +1 =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 value in the original equation f(9/2) = [9/2 - 4]^2+1= 1/4 +1 = 5/4 so, the coordinates of c (c,f(c)) is (9/2, 5/4) Mean Value Theorem for Derivatives states that if f(x) is a continuous function on [a,b] and differentiable on (a,b) then there exists a number c between a and b such that, f'(c)= [f(b)-f(a)]/(b-a) Mean Value Theorem for Integrals It states that if f(x) is a continuous function on [a,b], then there exists a number c in [a,b] such that, f(c)= 1/(b-a) [Integral (a to b)f(x) dx] This is the First Mean Value Theorem for Integrals From the theorem we can say that the average value of f on [a,b] is attained on [a,b].

Example: Let f(x) = 5x^4+2. Determine c, such that f(c) is the average value of f on the interval [-1,2] Solution: Using the Mean Value Theorem for the Integrals, f(c) = 1/(b-a)[integral(a to b) f(x) dx] The average value of f on the interval [-1,2] is given by, = 1/[2-(-1)] integral (-1 to 2) [5x^4+2]dx = 1/3 [x^5 +2x](-1 to 2) = 1/3 [ 2^5+ 2(2) -{(-1)^5+2(-1)}] = 1/3 [32+4+1+2] = 39/3 = 13 As f(c)= 5c^4+2, we get 5c^4+2 = 13, so c =+/-(11/5)^(1/4) We get, c= fourth root of (11/5) Second Mean Value Theorem for the integrals states that, If f(x) is continuous on an interval [a,b] then, d/dx Integral(a to b) f(t) dt = f(x) Example: find d/dx Integral (5 to x^2) sqrt(1+t^2)dt Solution: Applying the second Mean Value Theorem for Integrals, let u= x^2 which gives us y= integral (5 to u) sqrt(1+t^2)dt We know, dy/dx = dy/du. du.

dx = [sqrt(1+u^2)] (2x) = 2x[sqrt(1+x^4)]