Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be the function defined by
\[ f(x) = \begin{cases} x^2 - 4x - 5 & \text{if } x \ge 1 2x & \text{if } x < 1 \end{cases} \]
Which one of the following statements is TRUE?
Show Hint
If the ranges of two continuous parts of a piecewise function overlap heavily (here, $(-\infty, 2)$ and $[-9, \infty)$ overlap on $[-9, 2)$), the function will fail the horizontal line test and will not be one-one.
Step 1 : Understanding the Question:
The question asks us to analyze the given piecewise function $f(x)$ and determine whether it is one-one (injective) and/or onto (surjective). Step 2 : Key Formulas and Approach:
• One-One (Injective): A function is one-one if $f(a) = f(b) \implies a = b$. Geometrically, any horizontal line must cross the graph at most once.
• Onto (Surjective): A function is onto if its Range is equal to the Codomain ($\mathbb{R}$).
We will analyze the range and behavior of $f(x)$ on both intervals. Step 3 : Detailed Explanation:
Let us examine the two parts of the piecewise function:
• Part 1: $x < 1$
Here, $f(x) = 2x$.
Since $x < 1$, the values of $f(x)$ lie in the range:
\[ \text{Range}_1 = (-\infty, 2) \]
This function is strictly increasing since the derivative is $2 > 0$.
• Part 2: $x \ge 1$
Here, $f(x) = x^2 - 4x - 5$.
We can rewrite this by completing the square:
\[ f(x) = (x - 2)^2 - 9 \]
This is a parabola opening upwards with its vertex at $(2, -9)$.
Since the domain for this part is $x \ge 1$:
• At $x = 1$, $f(1) = 1 - 4 - 5 = -8$.
• At the vertex $x = 2$, it reaches its minimum value: $f(2) = -9$.
• As $x \to \infty$, $f(x) \to \infty$.
Thus, the range of this quadratic part is:
\[ \text{Range}_2 = [-9, \infty) \]
Let us evaluate onto-ness and one-one-ness:
• Onto-ness:
The total range of $f(x)$ is the union of the ranges of the two parts:
\[ \text{Range} = \text{Range}_1 \cup \text{Range}_2 = (-\infty, 2) \cup [-9, \infty) = (-\infty, \infty) = \mathbb{R} \]
Since the Range is equal to the Codomain ($\mathbb{R}$), the function is onto.
• One-one-ness:
Let us check if multiple values of $x$ map to the same value $y = 0$:
• From Part 1 ($x < 1$): $2x = 0 \implies x = 0$ (which is $< 1$, so it is a valid input).
• From Part 2 ($x \ge 1$): $x^2 - 4x - 5 = 0 \implies (x-5)(x+1) = 0 \implies x = 5$ (which is $\ge 1$, so it is a valid input).
Since $f(0) = 0$ and $f(5) = 0$ but $0 \neq 5$, the function is not one-one. Step 4 : Final Answer:
The function $f(x)$ is onto but not one-one.
This corresponds to Option (A).
