Math Problem Statement

solve this problem and draw me the graph. and show me the steps also

Sketch the graph and determine the domain and range of the following function: f(x) = {x + 9 if x < -2, |x - 2| + 1 if -2 ≤ x < 3, √(x - 2) if x > 3}.

Solution

Here is the graph of the piecewise function f(x)f(x) along with the steps to find the domain and range:

Domain:
The domain is (,)(-\infty, \infty), as the function is defined for all real numbers.

Range:
To determine the range, we analyze each part of the function:

  1. For x<2x < -2, f(x)=x+9f(x) = x + 9 is a linear function that decreases without bound as xx \to -\infty.
  2. For 2x<3-2 \leq x < 3, f(x)=x2+1f(x) = |x - 2| + 1. The minimum value occurs at x=2x = 2, where f(2)=1f(2) = 1, and it increases symmetrically away from this point.
  3. For x>3x > 3, f(x)=x2f(x) = \sqrt{x - 2}. The smallest value of this part is f(3)=1f(3) = 1, and as xx \to \infty, f(x)f(x) increases without bound.

Thus, the range of the function is (,)(-\infty, \infty), as the function covers all real values.

Would you like more details or have any questions?

Here are 5 related questions to further explore:

  1. How would the graph change if we modified the function to f(x)=x1+1f(x) = |x - 1| + 1 for the middle piece?
  2. What happens if we restrict the domain of the piecewise function to x>0x > 0?
  3. How does the graph behave if we use f(x)=(x+9)f(x) = -(x + 9) for x<2x < -2?
  4. Can you calculate the derivative of the piecewise function at x=2x = -2?
  5. What would happen if the square root function is replaced by (x2)2(x - 2)^2 for x>3x > 3?

Tip: For piecewise functions, it's important to carefully examine how each piece connects at the boundaries to ensure continuity or identify discontinuities.

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Math Problem Analysis

Mathematical Concepts

Piecewise Functions
Domain and Range
Absolute Value Functions
Square Root Functions

Formulas

f(x) = x + 9
f(x) = |x - 2| + 1
f(x) = √(x - 2)

Theorems

Properties of Piecewise Functions
Continuity and Discontinuity at Boundaries

Suitable Grade Level

Grades 9-12