Question

Let $f, g: \mathbb{R} \rightarrow \mathbb{R}$ be defined as: $f(x) = |x - 1|$ and $g(x) = \begin{cases} e^x, & x \geq 0 \\\ x + 1, & x \leq 0 \end{cases}$ Then the function $f(g(x))$ is

• A. neither one-one nor onto.
• B. one-one but not onto.
• C. both one-one and onto.
• D. onto but not one-one.

Answer 1

Answer: (A)

neither one-one nor onto.

Answer 2

To find $f(g(x))$, we first evaluate $g(x)$ based on the value of $x$.

$g(x) = \begin{cases} e^x, & x \geq 0 \\\ x + 1, & x \leq 0 \end{cases}$

The function $f$ is defined as $f(x) = |x - 1|$. Therefore, we have: $f(g(x)) = |g(x) - 1|.$

Case 1: When $x \geq 0$

$f(g(x)) = |e^x - 1|.$

Case 2: When $x \leq 0$

$f(g(x)) = |x + 1 - 1| = |x| = -x \quad \text{(since $ x \leq 0 $)}.$

Analysis of $f(g(x))$ - For $x \geq 0$, $f(g(x)) = |e^x - 1|$ is neither one-one nor onto because it cannot cover all values in the codomain (as it is non-negative). For $x \leq 0$, $f(g(x)) = -x$ is also neither one-one nor onto due to its behavior as a non-injective transformation on the interval.

Therefore, the function $f(g(x))$ is neither one-one nor onto.