Mean Value Theorem For Derivative. How do solve mean value theorem, how do I know a function is continue or not. Find out now

What is mean value theorem

The mean value theorem for derivatives if the function (F) is differenciable on open interval (a, b) and continues on close interval [a, b]. Then there exists at least C in between sacant line, and therefore a tangent line occur. see chain rule of partial derivative

Equation of Mean Value Theorem

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

Note:

[a, b] = Close interval

(a, b) = Open interval

Sacant line =[ f(b) - f(a) ]/b - a

Tangent line = F'(c)

Example 1

Find the value of C guaranteed by MVT for derivative for g(x) = (x + 1)³ on [-1, 1]

Soln

Since g(x) = (x + 1)³ on [a b] 

g'(x) = 3(x + 1)² × 1 

From the above expression the equation is continues and differentiable on the close interval and Open interval.

By MVT for derivative.

g'(c) = [g(b) - g(a)]/b - a.    eq1

Now  a = -1

          b = 1

g(b) = g(1) = (1 + 1)³ = (2)^3 = 8. eq2

g(a) = g(-1) = (-1 + 1)³ = 0.   eq3

By inserting eq1 and 2 into eq3 we have

(8 - 0)/(1- -1)

= 8/2

= 4

⇒3(c + 1)² = 4

⇒3(c + 1)(c + 1) = 4

⇒3(c² + 2c + 1) = 4

⇒3c² + 6c + 3 = 4

⇒3c² + 6c + 3 - 4 = 0

⇒3c² + 6c - 1 = 0 quadratic equation.

C = mighty formulae of quadratic equation.

Where 

a = 3, 

b = 6

c = -1

By using the formulae stated above we can now have 

C = 0.1547 ∈ [-1 1]

Exercises. Find the MVT

1) g(x) = (x - 4)/(x - 3) on interval of [0,5] or [4,6]

2) f(x) = cscx  on interval of [-π/2, π/2]

Roll's Theorem For mean value theorem

A diagram from roll's Theorem, for about MVT please click here