Determine whether the function $f(x) = \frac{1}{x}, x≥ 1$, is strictly increasing or decreasing for the stated values of x.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Strictly increasing for $x≥1$
Given $f(x) = \frac{1}{x}, x≥ 1⇒ f'(x) = 1-\frac{1}{x^2}$
Now $x≥1⇒x^2≥1⇒\frac{1}{x^2}≤1⇒0≤1-\frac{1}{x^2}⇒1-\frac{1}{x^2}≥0$
$∴f'(x) ≥ 0$ for $x ≥ 1$.
Note that $f'(x) = 0$ at $x = 1$ and $f'(x) > 0$ for all $x > 1$.
Therefore, f(x) is strictly increasing for $x ≥ 1$.