Question

Verify mean value theorem for function 3x^2-5x+1 defined in interval [2,5]

Answer 1

Mean Value Theorem :

Let f(x) be a real valued function that satisfies the following conditions.

(i) f(x) is continuous on the closed interval [a,b]

(ii) f(x) is differentiable in the open interval (a,b)

Then there exists atleast one value c∈(a,b) such that ,

f′(c)=f(b)−f(a)/(b−a)

Verify mean value theorem for the following function:

Step 1:

Given :f(x)=3x^2−5x+1 in the interval [2,5]

We know that a polynomial function is continuous everywhere and also differentiable.

So f(x) being a polynomial is continuous and differentiable on (2,5)

So there must exist at least one real number c∈(2,5) such that

f′(c)=f(2)−f(5)/(2-5)

Step 2:

f(x)=3x^2−5x+1

f(2)=3

f(5)=51

f′(x)=6x−5

f′(c)=6c−5

∴6c−5=3-51/(2-5)

6c−5=-48/-3

c=3.5

c∈(2,5)

Hence LMean Value theorem is verified.