Given that f is continuous and differentiable in the interval [ – 5,5].
It is also given that f’(x) doesn’t vanish anywhere.
According to Rolle’s theorem for a differentiable function on [a,b] will have atleast one cϵ(a,b) such that f’(c) = 0, if the following condition had satisfied:
⇒ f(a) = f(b).
According to the problem it is given for any value of x, say r the values never equals to zero.
This is possible when Rolle’s theorem is not applicable.
Let us Recap the Rolle’s theorem:
For a Real valued function ‘f’:
a) The function ‘f’ needs to be continuous in the closed interval [a,b].
b) The function ‘f’ needs differentiable on the open interval (a,b).
c) f(a) = f(b)
Then there exists at least one c in the open interval (a,b) such that f’(c) = 0.
First, two conditions are satisfied according to the problem, so the only condition that cannot be satisfied is (c).
So, we can clearly say that f( – 5)≠f(5).
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