Question

Discuss the applicability of Rolle’s theorem on the function given by f(x) = `{{:(x^2 + 1",",  "if"  0 ≤ x ≤ 1),(3 - x",",  "if"  1 ≤ x ≤ 2):}`

Answer 1

We have, f(x) = `{{:(x^2 + 1",",  "if"  0 ≤ x ≤ 1),(3 - x",",  "if"  1 ≤ x ≤ 2):}`

We know that polynomial function is everywhere continuous and differentiability.

So, f(x) is continuous and differentiable at all points except possibly at x = 1.

Now `lim_(x -> 1^-) (x^2 + 1)` = 1 + 1 = 2

And `lim_(x -> 1^+) (3 - x)` = 3 – 1 = 2

Also f(1) = 1² + 1 = 2

So, f(x) is continuous at x = 1

Also f'(x) = `{{:(2x",",  "if"  0 < x < 1),(-x",",  "if"  1 < x 2):}`

f'(1) = 2(1) = 2

And f'(1) = –1

Thus f'(1) ≠ f'(1).

So, f(x) is not differentiable at x = 1

Hence, Rolle's theorem is not applicable on the interval [0, 2].