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].
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).
Here, f(1)= (1-1)(1-2)(1-3) =0
And f(3)= (3-1)(3-2)(3-3) =0
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=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 :
The value of c inMathematics - Exemplar
For the functionMathematics - Exemplar
Discuss theRD Sharma - Volume 1
Using Rolle’s theMathematics - Exemplar
State TrueMathematics - Exemplar
Discuss the appliMathematics - Exemplar