(B) Given the function $f(x) = \begin{cases} 0, & x = 0 \\ x - 3, & x > 0 \end{cases}$.For $x > 0$,the derivative is $f'(x) = \frac{d}{dx}(x - 3) = 1$

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$A$ strictly increasing function for $x > 0$.

(B) Given the function $f(x) = \begin{cases} 0, & x = 0 \\ x - 3, & x > 0 \end{cases}$.For $x > 0$,the derivative is $f'(x) = \frac{d}{dx}(x - 3) = 1$.Since $f'(x) = 1 > 0$ for all $x > 0$,the function $f(x)$ is strictly increasing on the interval $(0, \infty)$.Therefore,$f(x)$ is a strictly increasing function for $x > 0$.

Class 12 Mathematics Applications of Derivatives Increasing and Decreasing function

Medium

If $f(x) = \begin{cases} 0, & x = 0 \\ x - 3, & x > 0 \end{cases}$,then $f(x)$ is:

A
An increasing function for $x \geq 0$.
B
$A$ strictly increasing function for $x > 0$.
C
$A$ strictly decreasing function at $x = 0$.
D
Not continuous at $x = 0$,therefore it is not an increasing function for $x > 0$.