Chapter 1 Functions And Their Graphs 1.2 Exercises Answers Precalculus

understandingfunctions and their graphs is fundamental to precalculus, forming the bedrock for more advanced mathematical concepts. this chapter delves into defining functions, evaluating them, and interpreting their graphical representations, which are crucial for visualizing relationships between variables. mastering these skills enables students to analyze real-world phenomena, solve complex problems, and build a solid foundation for calculus. the exercises provided in section 1.2 offer essential practice in applying these core principles, reinforcing comprehension and analytical abilities.

1.2 exercises answers precalculus

the following solutions provide detailed explanations for the exercises in section 1.2, focusing on function evaluation, domain, range, and basic graph interpretation. each solution emphasizes the underlying concepts to deepen understanding.

exercise 1.1

1. evaluate the function f(x) = 2x + 3 at x = 4.

   • solution: substitute x = 4 into the function: f(4) = 2*(4) + 3 = 8 + 3 = 11. thus, f(4) = 11.
   • explanation: evaluating a function means finding the output (y-value) corresponding to a specific input (x-value). here, the linear function f(x) = 2x + 3 has a slope of 2 and y-intercept of 3. plugging in x=4 gives the point (4, 11) on the graph.

2. find the domain and range of the function g(x) = √(x - 2).

   • solution: the domain is x ≥ 2 (since the expression under the square root must be non-negative). the range is y ≥ 0 (as square roots yield non-negative results).
   • explanation: the domain represents all possible x-values that produce a real output. the range represents all possible y-values produced. for a square root function, the domain restriction comes from requiring the radicand (x-2) ≥ 0.

3. determine if the relation {(1, 3), (2, 4), (3, 5), (1, 2)} is a function. explain.

   • solution: no, it is not a function. the input x=1 has two different outputs: y=3 and y=2.
   • explanation: a function must assign exactly one output to each input. here, the input 1 maps to two different outputs, violating the definition of a function.

exercise 1.2

1. evaluate the function h(x) = x² - 4x + 3 at x = 0, x = 1, and x = 3.

   • solution:
     • h(0) = (0)² - 4*(0) + 3 = 0 - 0 + 3 = 3.
     • h(1) = (1)² - 4*(1) + 3 = 1 - 4 + 3 = 0.
     • h(3) = (3)² - 4*(3) + 3 = 9 - 12 + 3 = 0.
   • explanation: evaluating a quadratic function involves substituting the given x-value into the expression and simplifying. the results give the corresponding y-values (h-values) for each input.

2. find the domain and range of the function k(x) = 1 / (x - 5).

   • solution: the domain is x ≠ 5 (all real numbers except 5). the range is y ≠ 0 (all real numbers except zero).
   • explanation: division by zero is undefined, so x=5 is excluded from the domain. the function never reaches y=0 because the numerator is constantly 1. as x approaches 5 from either side, k(x) becomes very large positive or negative, and as x moves far away, k(x) approaches zero.

3. graph the function m(x) = |x - 3|. label the vertex and state the domain and range.

   • solution: the graph is a "v" shape with its vertex at (3, 0). the domain is all real numbers (x ∈ ℝ). the range is y ≥ 0 (y ∈ [0, ∞)).
   • explanation: the absolute value function |x - 3| represents the distance between x and 3 on the number line. it has a sharp turn (vertex) at x=3 where the expression inside the absolute value equals zero. the graph is symmetric about the line x=3.

4. determine the domain and range of the function n(x) = 2x - 7 for x in the interval [1, 4].

   • solution: the domain is [1, 4]. the range is [21 - 7, 24 - 7] = [-5, 1].
   • explanation: the domain is restricted to the given interval [1, 4]. since n(x) is linear, the range is found by evaluating the function at the endpoints: n(1) = -5 and n(4) = 1. the range is the interval between these values, including both endpoints.

5. evaluate the piecewise function p(x) = { 3x + 1 if x < 2, x² if x ≥ 2 } at x = 1, x = 2, and x = 3.

   • solution:
     • p(1) = 3*(1) + 1 = 4 (since 1 < 2).
     • p(2) = (2)² = 4 (since 2 ≥ 2).
     • p(3) = (3)² = 9 (since 3 ≥

2).

• Explanation: For a piecewise function, the correct formula is chosen based on the input value. For x = 1, since 1 < 2, we use the first formula (3x + 1). For x = 2 and x = 3, since both are greater than or equal to 2, we use the second formula (x²).

Exercise 1.3

1. Determine whether the relation {(2, 4), (3, 9), (4, 16), (5, 25)} is a function. Explain.

   • Solution: Yes, it is a function.
   • Explanation: Each input (x-value) has exactly one output (y-value). There are no repeated x-values with different y-values.

2. Determine whether the relation {(1, 2), (2, 4), (1, 3), (3, 6)} is a function. Explain.

   • Solution: No, it is not a function.
   • Explanation: The input x = 1 has two different outputs: y = 2 and y = 3. A function must assign exactly one output to each input.

3. Find the domain and range of the function q(x) = √(x + 4).

   • Solution: The domain is x ≥ -4 (x ∈ [-4, ∞)). The range is y ≥ 0 (y ∈ [0, ∞)).
   • Explanation: The expression under the square root must be non-negative, so x + 4 ≥ 0, which means x ≥ -4. The square root of a non-negative number is always non-negative, so the range is all non-negative real numbers.

4. Graph the function r(x) = -2x + 5. Label the y-intercept and state the domain and range.

   • Solution: The graph is a straight line with a y-intercept at (0, 5). The domain is all real numbers (x ∈ ℝ). The range is all real numbers (y ∈ ℝ).
   • Explanation: This is a linear function in slope-intercept form (y = mx + b). The y-intercept is the point where x = 0, which is (0, 5). Since it's a non-vertical line, both the domain and range include all real numbers.

5. Evaluate the function s(x) = x³ - 2x + 1 at x = -1, x = 0, and x = 2.

   • Solution:
     • s(-1) = (-1)³ - 2*(-1) + 1 = -1 + 2 + 1 = 2.
     • s(0) = (0)³ - 2*(0) + 1 = 0 - 0 + 1 = 1.
     • s(2) = (2)³ - 2*(2) + 1 = 8 - 4 + 1 = 5.
   • Explanation: Evaluating a cubic function involves substituting the given x-value into the expression and simplifying. The results give the corresponding y-values (s-values) for each input.

Conclusion

Understanding functions is fundamental to algebra and higher mathematics. A function is a rule that assigns exactly one output to each input, and it can be represented in various ways, including equations, graphs, tables, and mappings. Key concepts include the domain (set of all possible inputs), range (set of all possible outputs), and the vertical line test (to determine if a graph represents a function). Evaluating functions involves substituting input values into the function's rule and simplifying. Piecewise functions use different rules for different intervals of the domain. Mastery of these concepts is essential for solving equations, analyzing relationships between variables, and exploring more advanced mathematical topics.