Show that the function x2 – x + 1 is neither increasing nor decreas

1.	Show that the function x2 – x + 1 is neither increasing nor decreasing on (0, 1).
Answer» Given as f(x) = x² – x + 1 Differentiate the given equation with respect to x, we get ⇒ f'(x) = (d/dx)(x² - x + 1) ⇒ f’(x) = 2x – 1 On taking different region from (0, 1) Suppose x ∈ (0, 1/2) ⇒ 2x – 1 < 0 ⇒ f’(x) < 0 Hence f(x) is decreasing in (0, 1/2) Suppose x ∈ (1/2, 1) ⇒ 2x – 1 > 0 ⇒ f’(x) > 0 Hence f(x) is increasing in (1/2, 1) So, from above condition we find that ⇒ f (x) is decreasing in (0, 1/2) and increasing in (1/2, 1) Thus, condition for f(x) neither increasing nor decreasing in (0, 1)

Show that the function x2 – x + 1 is neither increasing nor decreasing on (0, 1).

Answer»

Given as f(x) = x² – x + 1

Differentiate the given equation with respect to x, we get

⇒ f'(x) = (d/dx)(x² - x + 1)

⇒ f’(x) = 2x – 1

On taking different region from (0, 1)

Suppose x ∈ (0, 1/2)

⇒ 2x – 1 < 0

⇒ f’(x) < 0

Hence f(x) is decreasing in (0, 1/2)

Suppose x ∈ (1/2, 1)

⇒ 2x – 1 > 0

⇒ f’(x) > 0

Hence f(x) is increasing in (1/2, 1)

So, from above condition we find that

⇒ f (x) is decreasing in (0, 1/2) and increasing in (1/2, 1)

Thus, condition for f(x) neither increasing nor decreasing in (0, 1)