Math Problem Statement

Quesripn 10

Determine the formula for the piecewise-defined function in the graph, find its domain and range, and calculate specific function values.

Solution

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Question 10:

  • A function f(x)f(x) is defined with two cases: \begin{cases} x - 1 & \text{if } x < 2 \\ 2 - x^2 & \text{if } x \geq 2 \end{cases}$$
  1. Graph the function f(x)f(x).
  2. Find the domain and range of f(x)f(x).
  3. Calculate f(1)f(-1) and f(1+3)f(1 + \sqrt{3}).

Solutions:

1. Graph the Function f(x)f(x):

  • Case 1: When x<2x < 2, f(x)=x1f(x) = x - 1, which is a linear function with a slope of 1 and a y-intercept of -1.
  • Case 2: When x2x \geq 2, f(x)=2x2f(x) = 2 - x^2, which is a downward parabola with a vertex at (0,2)(0, 2).

By combining these pieces, we can draw the function with the following points and conditions:

  • For x<2x < 2, graph the line y=x1y = x - 1.
  • For x2x \geq 2, graph the parabola y=2x2y = 2 - x^2.

2. Domain and Range:

  • Domain: The function is defined for all real numbers xx. Hence, the domain is: Domain=(,)\text{Domain} = (-\infty, \infty)

  • Range: The line y=x1y = x - 1 for x<2x < 2 covers (,1)(-\infty, 1). The parabola y=2x2y = 2 - x^2 for x2x \geq 2 has a vertex at (2,2)(2, -2) and decreases infinitely. Thus, the overall range is: Range=(,1)[2,2]\text{Range} = (-\infty, 1) \cup [-2, 2]

3. Calculate f(1)f(-1) and f(1+3)f(1 + \sqrt{3}):

  • f(1)f(-1): Since 1<2-1 < 2, use the formula f(x)=x1f(x) = x - 1: f(1)=11=2f(-1) = -1 - 1 = -2

  • f(1+3)f(1 + \sqrt{3}): Since 1+32.7321 + \sqrt{3} \approx 2.732 and 2.732>22.732 > 2, use the formula f(x)=2x2f(x) = 2 - x^2: f(1+3)=2(1+3)2f(1 + \sqrt{3}) = 2 - (1 + \sqrt{3})^2 Expand (1+3)2(1 + \sqrt{3})^2: (1+3)2=1+23+3=4+23(1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} So: f(1+3)=2(4+23)=223f(1 + \sqrt{3}) = 2 - (4 + 2\sqrt{3}) = -2 - 2\sqrt{3}


Final Results:

  1. Graph: Constructed by combining the line y=x1y = x - 1 for x<2x < 2 and the parabola y=2x2y = 2 - x^2 for x2x \geq 2.
  2. Domain: (,)(-\infty, \infty).
  3. Range: (,1)[2,2](-\infty, 1) \cup [-2, 2].
  4. f(1)=2f(-1) = -2.
  5. f(1+3)=223f(1 + \sqrt{3}) = -2 - 2\sqrt{3}.

Would you like me to graph the function or explain any specific part in more detail?


Related Questions:

  1. What is the difference between piecewise-defined functions and continuous functions?
  2. How do you determine the range of a quadratic function graphically?
  3. What is the effect of x<cx < c and xcx \geq c conditions in piecewise functions?
  4. How do you calculate function values for irrational arguments like 3\sqrt{3}?
  5. How would the graph and range change if f(x)f(x) were modified to f(x)=x22x+1f(x) = x^2 - 2x + 1 for x<2x < 2?

Tip:

Always carefully evaluate the conditions for piecewise functions to ensure the correct formula is applied!

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Math Problem Analysis

Mathematical Concepts

Piecewise Functions
Linear Equations
Quadratic Functions
Domain and Range

Formulas

Piecewise Function Definition: f(x) = {expression_1 for condition_1, expression_2 for condition_2}
Linear Function: y = mx + b
Quadratic Function: y = ax^2 + bx + c

Theorems

Domain and Range Analysis
Evaluating Piecewise Functions

Suitable Grade Level

Grades 10-12