Question

Prove that the function f given by f(x) = x^2 − x + 1 is neither strictly increasing nor strictly decreasing on (−1, 1).

Answer 1

HELLO DEAR,

f(x) = (x² - x + 1)

f'(x) = (2x - 1).

now, f'(x) > 0

(2x - 1) > 0

x > 1/2.

therefore, f(x) is increasing when x € ]1/2 , 1[.

and, f'(x) < 0

(2x - 1) < 0

x < 1/2.

therefore, f(x) is deacreasing when x € ]0 , 1/2[.


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

given, function f(x) = x² – x + 1

f’(x) = 2x - 1

If f’(x) = 0, then we get,

x = 1/2

here, the point x = 1/2 divides the interval (-1,1) into two disjoint intervals (-1, 1/2) and [1/2, 1)

Let's take interval (-1, 1/2)

we get , f'(x) = 2x - 1 < 0

therefore, f is decreasing in interval (-1,1/2)

Let's take interval [1/2, 1)

we get, f'(x) = 2x - 1 > 0

therefore , f is increasing in interval [1/2, 1)

Therefore, f is neither strictly increasing and decreasing in interval (-1,1).