Common Graphs — Question 3

Question 3 - Solution

We are given a piecewise function: $$f(x) = \begin{cases} x^2 + 1 & \text{if } x < 0 \\ 2x + 1 & \text{if } x \geq 0 \end{cases}$$

Step 1: Domain

Both pieces are defined on their respective intervals, so: $$\boxed{\text{Domain: } (-\infty, \infty)}$$

Step 2: Range

For $x < 0$: $f(x) = x^2 + 1 \Rightarrow f(x) > 1$ , For $x \geq 0$: $f(x) = 2x + 1 \Rightarrow f(x) \geq 1$

So the function reaches its minimum at $f(0) = 1$, and increases in both directions: $$\boxed{\text{Range: } [1, \infty)}$$

Step 3: Continuity at $x = 0$

$$\lim_{x \to 0^-} f(x) = 0^2 + 1 = 1, \quad \lim_{x \to 0^+} f(x) = 2(0) + 1 = 1 \Rightarrow \text{Function is continuous at } x = 0$$

Step 4: Intercepts

Y-intercept: $f(0) = 1$ X-intercept: Solve $f(x) = 0$: , $x^2 + 1 = 0 \Rightarrow \text{no real solution}$ , $2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \notin [0, \infty) \Rightarrow \text{no x-intercept}$

Conclusion: Domain: $(-\infty, \infty)$ , Range: $[1, \infty)$ , No x-intercepts, y-intercept at $(0, 1)$ , Continuous everywhere