Question

Which of the following function are monotonic in the interval (0,1) ?

• A. $f\left(x\right)=x^2$
• B. $f\left(x\right)=\frac{1}{x}$
• C. $f\left(x\right)=x^3-x$
• D. $f\left(x\right)=ln\left(x\right)$

Answer 1

Answer: (A), (B), (D)

$f\left(x\right)=ln\left(x\right)$

To figure out the intervals in which a function is monotonous we’ll differentiate the function and find the intervals in which the derivative has the same sign

Function will be monotonic when f’(x) > 0 or f’(x) < 0 or f’(x) $\ge$ 0 or f’(x) $\le$ 0.

Let’s differentiate each option one by one.

1. f(x) = $x^2$

In the interval (0,1) f’(x) > 0

So we can say f(x) = $x^2$ is monotonic in (0,1)

B. f(x) = 1/x

f’(x) = -1 / $x^2$

Here, we can see that f’(x) $\le$ 0 in the interval (0,1)

So the function is monotonically increasing.

C. f(x) = $x^3$- x

f’(x) = 3 $x^2$-1

f’(x) will be positive in the intervals of -

3 $x^2$-1 ≥ 0

$x^2$- 1/3 ≥ 0

$\Rightarrow xϵ(-\infty ,-1/\sqrt{3}]U[1/\sqrt{3},\infty )ϵ$

And $x^2-\frac{1}{3}$ will be negative in the interval [- 1 / $\sqrt{3}$, 1 / $\sqrt{3}$]

In the interval (0,1)for one part of the interval that is (0, 1 / $\sqrt{3}$ ] function will be decreasing and in [1 / $\sqrt{3}$ , 1) it’ll be increasing.

Hence, the function is not monotonic in the interval (0,1).

D. f(x) = ln(x)

f’(x) = 1/x

f’(x) is positive in the interval of (0,1) and thus monotonic.