Math Problem Statement
The function f is defined as follows. f left parenthesis x right parenthesis equalsleft brace Start 2 By 1 Matrix 1st Row 1st Column 3 plus x 2nd Row 1st Column x squared EndMatrix Start 2 By 1 Matrix 1st Row 1st Column if x less than 0 2nd Row 1st Column if x greater than or equals 0 EndMatrix
(a) Find the domain of the function. (b) Locate any intercepts. (c) Graph the function. (d) Based on the graph, find the range.
Solution
We are given a piecewise function , defined as:
\begin{cases} 3 + x & \text{if } x < 0, \\ x^2 & \text{if } x \geq 0. \end{cases}$$ Let's go through each part of the question. ### (a) Domain of the Function The domain of a function is the set of all possible input values (x-values) for which the function is defined. - For $$x < 0$$, $$f(x) = 3 + x$$, which is defined for all negative values of $$x$$. - For $$x \geq 0$$, $$f(x) = x^2$$, which is defined for all non-negative values of $$x$$. Since there are no restrictions on $$x$$ in either part of the piecewise function, the domain of $$f(x)$$ is all real numbers. **Domain:** $$(-\infty, \infty)$$ ### (b) Intercepts **Intercepts** are where the function crosses the x-axis (x-intercepts) or the y-axis (y-intercept). 1. **Y-Intercept:** The y-intercept occurs when $$x = 0$$. - Since $$f(x) = x^2$$ when $$x \geq 0$$, we evaluate: $$f(0) = 0^2 = 0.$$ So, the y-intercept is $$(0, 0)$$. 2. **X-Intercepts:** X-intercepts occur when $$f(x) = 0$$. - For $$x < 0$$, $$f(x) = 3 + x$$. Set $$3 + x = 0$$: $$x = -3.$$ So, there is an x-intercept at $$(-3, 0)$$. - For $$x \geq 0$$, $$f(x) = x^2$$. Set $$x^2 = 0$$: $$x = 0.$$ So, there is another x-intercept at $$(0, 0)$$, but this is already the y-intercept. **Intercepts:** $$(-3, 0)$$ and $$(0, 0)$$ ### (c) Graph of the Function To graph the function, we consider the behavior of the two pieces: - For $$x < 0$$, the function is a linear equation $$f(x) = 3 + x$$, with slope 1 and y-intercept 3. - For $$x \geq 0$$, the function is a parabola $$f(x) = x^2$$, which opens upwards with its vertex at $$(0, 0)$$. The two pieces meet at $$x = 0$$, where $$f(0) = 0$$. We can sketch the graph by plotting key points: - For $$x < 0$$, plot points like $$(-1, 2)$$, $$(-2, 1)$$, and $$(-3, 0)$$. - For $$x \geq 0$$, plot points like $$(1, 1)$$, $$(2, 4)$$, and so on. Let me know if you want me to plot this for you. ### (d) Range of the Function The **range** of the function is the set of all possible output values (y-values) the function can take. - For $$x < 0$$, $$f(x) = 3 + x$$, which is a linear function with values approaching 3 as $$x \to 0$$ from the left, and continuing downward as $$x \to -\infty$$. - For $$x \geq 0$$, $$f(x) = x^2$$, which is a parabola that starts at $$f(0) = 0$$ and increases to infinity as $$x \to \infty$$. So, the function covers all values from $$-\infty$$ to 3 in the negative region, and all values from 0 to $$\infty$$ in the non-negative region. The overall range of the function is: **Range:** $$(-\infty, \infty)$$ --- Would you like more details on any part, or a graph? Here are some related questions: 1. What would happen to the intercepts if the function were modified? 2. How would the graph change if $$f(x) = x^3$$ for $$x \geq 0$$? 3. Can the range change for a similar piecewise function if one piece is exponential? 4. How do the continuity and differentiability of this function behave at $$x = 0$$? 5. What is the slope of the linear part of $$f(x)$$ for $$x < 0$$? **Tip:** The continuity of a piecewise function is essential to consider at the points where the definition changes.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Domain and Range
Graphing Functions
Intercepts
Formulas
Piecewise function definitions: f(x) = {3 + x, if x < 0; x^2, if x >= 0}
Theorems
Concepts of domain and range
Properties of linear and quadratic functions
Suitable Grade Level
Grades 9-12