Answer :

Condition (1):

Since, f(x)=(x-1)(x-2)(x-3) is a polynomial and we know every polynomial function is continuous for all xϵR.

f(x)= (x-1)(x-2)(x-3) is continuous on [1,3].

Condition (2):

Here, f’(x)= (x-2)(x-3)+ (x-1)(x-3)+ (x-1)(x-2) which exist in [1,3].

So, f(x)= (x-1)(x-2)(x-3) is differentiable on (1,3).

Condition (3):

Here, f(1)= (1-1)(1-2)(1-3) =0

And f(3)= (3-1)(3-2)(3-3) =0

i.e. f(1)=f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(1,3) such that f’(c)=0

i.e. (c-2)(c-3)+ (c-1)(c-3)+ (c-1)(c-2)=0

i.e. (c-3)(2c-3)+(c-1)(c-2)=0

i.e. (2c2-9c+9)+(c2-3c+2)=0

i.e. 3c2-12c+11=0


i.e. c=2.58 or c=1.42

Value of c=1.42ϵ(1,3) and c=2.58ϵ(1,3)

Thus, Rolle’s theorem is satisfied.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

Verify the Rolle’Mathematics - Exemplar

The value of c inMathematics - Exemplar

For the function Mathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

Discuss theRD Sharma - Volume 1

Using Rolle’s theMathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

State True Mathematics - Exemplar

Discuss the appliMathematics - Exemplar