Show that the function x2 – x + 1 is neither increasing nor decreasing on (0, 1).
Given:- Function f(x) = x2 – x + 1
Theorem:- Let f be a differentiable real function defined on an open interval (a,b).
(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)
(i) Obtain the function and put it equal to f(x)
(ii) Find f’(x)
(iii) Put f’(x) > 0 and solve this inequation.
For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.
Here we have,
f(x) = x2 – x + 1
⇒ f’(x) = 2x – 1
Taking different region from (0, 1)
⇒ 2x – 1 < 0
⇒ f’(x) < 0
Thus f(x) is decreasing in
⇒ 2x – 1 > 0
⇒ f’(x) > 0
Thus f(x) is increasing in
Therefore, from above condition we find that
⇒ f(x) is decreasing in and increasing in
Hence, condition for f(x) neither increasing nor decreasing in (0, 1)