Math Problem Statement
Consider the following piecewise function: f(x) = { x^2 + 2x, if x ≤ 0; 2x - 1, if x > 0. Determine the domain and range of the function, and sketch its graph.
Solution
Let's analyze the given problem step by step.
The piecewise function is:
\begin{cases} x^2 + 2x, & \text{if } x \leq 0, \\ 2x - 1, & \text{if } x > 0. \end{cases}$$ We need to address two parts: --- ### (a) **Domain and Range of the Function** #### **Domain** The domain of a piecewise function is the union of the domains of the individual pieces. Both pieces are polynomials, which are defined for all real numbers: - For $$x^2 + 2x$$, valid for $$x \leq 0$$, - For $$2x - 1$$, valid for $$x > 0$$. Thus, the **domain** of $$f(x)$$ is: $$\text{Domain: } (-\infty, \infty).$$ --- #### **Range** To find the range, we evaluate the outputs of the function for both pieces and check the continuity at the transition point ($$x = 0$$): 1. **For $$x \leq 0$$:** The expression $$x^2 + 2x$$ is a quadratic function opening upwards ($$a > 0$$) with a vertex at: $$x_{\text{vertex}} = -\frac{b}{2a} = -\frac{2}{2(1)} = -1.$$ Substituting $$x = -1$$ into $$f(x)$$, we get: $$f(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1.$$ So, for $$x \leq 0$$, the function decreases to $$-1$$ at $$x = -1$$ and then increases without bound as $$x \to 0^-$$. Thus, for $$x \leq 0$$, the range is $$[ -1, \infty )$$. 2. **For $$x > 0$$:** The expression $$2x - 1$$ is a linear function with a slope of 2. As $$x > 0$$, the values of $$f(x)$$ start from: $$f(0^+) = 2(0) - 1 = -1,$$ and increase without bound as $$x \to \infty$$. Thus, for $$x > 0$$, the range is $$(-1, \infty)$$. 3. **Overall Range:** Combining the two parts, the range of $$f(x)$$ is: $$\text{Range: } [ -1, \infty ).$$ --- ### (b) **Sketch of the Graph** #### Steps to Sketch: 1. **For $$x \leq 0$$:** Plot the parabola $$y = x^2 + 2x$$, focusing on the interval $$x \leq 0$$. It has a vertex at $$(-1, -1)$$ and passes through $$(0, 0)$$. 2. **For $$x > 0$$:** Plot the line $$y = 2x - 1$$, starting at the point $$(0, -1)$$ (not included) and increasing for $$x > 0$$. 3. **Verify Continuity:** At $$x = 0$$, the left-hand limit (from $$x \leq 0$$) is $$f(0) = 0^2 + 2(0) = 0$$, while the right-hand limit (from $$x > 0$$) is $$f(0^+) = 2(0) - 1 = -1$$. Since these do not match, there is a jump discontinuity at $$x = 0$$. Would you like me to graph this function for you? Let me know! --- ### Questions to Expand Your Understanding: 1. How do we determine continuity for a piecewise function at the transition point? 2. What is the significance of the vertex in determining the range for a quadratic piece of the function? 3. How would the domain or range change if one of the function pieces were defined for a stricter condition (e.g., $$x < 0$$ instead of $$x \leq 0$$)? 4. How can we determine if a piecewise function is differentiable at the transition point? 5. What are other examples of piecewise functions that combine linear and quadratic components? --- ### Tip: When analyzing piecewise functions, always check continuity at transition points by comparing the left-hand and right-hand limits.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Quadratic Equations
Linear Equations
Domain and Range
Continuity
Formulas
Vertex formula for quadratics: x_vertex = -b/(2a)
Quadratic equation: ax^2 + bx + c
Linear equation: y = mx + b
Theorems
Properties of Quadratic Functions
Properties of Linear Functions
Continuity and Limits in Piecewise Functions
Suitable Grade Level
Grades 10-12