Answered

  1. Find the domain and range of each of the following relations:
A$R_1 = \{(1,0), (2,2), (3,2), (4,5)\}$
B$R_2 = \{(-1,-2), (0,1), (1,2), (2,5)\}$
C$R_3 = \{(x, y) \mid x, y \in \mathbb{R} \text{ and } y = x - 1\}$
D$R_4 = \{(x, y) \mid x, y \in \mathbb{R} \text{ and } y = x^2 + 1\}$


  1. Find the inverse, domain, and range of the inverses of each of the relations in Question 1 above.


  1. Find the inverse of each of the following functions:
A$f(x) = 3x - 2$
B$g(x) = x^3 + 5$
C$h(x) = \frac{2x+3}{4x-5}$


  1. Find the domain and range of each of the following functions:
A$f(x) = 3x^{\frac{4}{5}}$
B$g(x) = -2x^{\frac{5}{4}}$
C$h(x) = x^{\frac{3}{5}}$


  1. Given $f(x) = 3 + 2|x|$:
AEvaluate $f$ at each of the following numbers:

i. $-5$

ii. $0$

iii. $-\frac{5}{2}$

iv. $3$

BFind the domain and range of $f$.
CSketch the graph of $f$.


  1. Given $f(x) = 2 - 3\operatorname{sgn}(x)$:

Evaluate $f$ at each of the following numbers.

Verified Answer :

  • For Q1 and Q2, identify domains and ranges by listing coordinates and swapping them for inverses.
  • For Q3, solve for \(x\) in terms of \(y\) to find inverse functions: \(f^{-1}(x) = \frac{x+2}{3}\), \(g^{-1}(x) = \sqrt[3]{x-5}\), \(h^{-1}(x) = \frac{5x+3}{4x-2}\).
  • For Q4 and Q5, determine domain and range by analyzing properties like \(\sqrt[n]{x}\) and \(|x| \ge 0\).
  • For Q6, use the definition of the signum function to evaluate the function for given inputs.

Explanation

  1. Analysis of Relations (Q1)
  2. Question 1: Relations
  • \(R_1 = \{(1,0), (2,2), (3,2), (4,5)\}\):
  • Domain: Set of first coordinates, \(\{1, 2, 3, 4\}\).
  • Range: Set of second coordinates, \(\{0, 2, 5\}\).
  • \(R_2 = \{(-1,-2), (0,1), (1,2), (2,5)\}\):
  • Domain: \(\{-1, 0, 1, 2\}\).
  • Range: \(\{-2, 1, 2, 5\}\).
  • \(R_3 = \{(x, y) \mid y=x-1\}\):
  • Domain: Any real \(x\), \((-\infty, \infty)\).
  • Range: Any real \(y\), \((-\infty, \infty)\).
  • \(R_4 = \{(x, y) \mid y=x^2+1\}\):
  • Domain: Any real \(x\), \((-\infty, \infty)\).
  • Range: Since \(x^2 \ge 0\), \(y = x^2+1 \ge 1\), \([1, \infty)\).
  1. Analysis of Inverse Relations (Q2)
  2. Question 2: Inverse Relations
  • \(R_1^{-1} = \{(0,1), (2,2), (2,3), (5,4)\}\); Domain: \(\{0, 2, 5\}\), Range: \(\{1, 2, 3, 4\}\).
  • \(R_2^{-1} = \{(-2,-1), (1,0), (2,1), (5,2)\}\); Domain: \(\{-2, 1, 2, 5\}\), Range: \(\{-1, 0, 1, 2\}\).
  • \(R_3^{-1}\): For \(y=x-1\), the inverse is \(x=y-1 \implies y=x+1\). Domain: \((-\infty, \infty)\), Range: \((-\infty, \infty)\).
  • \(R_4^{-1}\): For \(y=x^2+1\), \(x^2=y-1 \implies x=\pm\sqrt{y-1}\). Domain: \([1, \infty)\), Range: \((-\infty, \infty)\).
  1. Analysis of Inverse Functions (Q3)
  2. Question 3: Inverse Functions
  • a. \(f(x) = 3x - 2 \implies y = 3x-2 \implies x = \frac{y+2}{3} \implies f^{-1}(x) = \frac{x+2}{3}\).
  • b. \(g(x) = x^3 + 5 \implies y = x^3+5 \implies x = \sqrt[3]{y-5} \implies g^{-1}(x) = \sqrt[3]{x-5}\).
  • c. \(h(x) = \frac{2x+3}{4x-5} \implies y(4x-5) = 2x+3 \implies 4xy-5y = 2x+3 \implies x(4y-2) = 5y+3 \implies x = \frac{5y+3}{4y-2} \implies h^{-1}(x) = \frac{5x+3}{4x-2}\).
  1. Analysis of Power Functions (Q4)
  2. Question 4: Domain and Range of Power Functions
  • a. \(f(x) = 3x^{4/5} = 3\sqrt[5]{x^4}\):
  • Domain: All real \(x\), \((-\infty, \infty)\).
  • Range: \(\sqrt[5]{x^4} \ge 0\), so Range is \([0, \infty)\).
  • b. \(g(x) = -2x^{5/4} = -2\sqrt[4]{x^5}\):
  • Domain: Requires \(x \ge 0\), \([0, \infty)\).
  • Range: \(x^{5/4} \ge 0 \implies -2x^{5/4} \le 0\), so Range is \((-\infty, 0]\).
  • c. \(h(x) = x^{3/5} = \sqrt[5]{x^3}\):
  • Domain: All real \(x\), \((-\infty, \infty)\).
  • Range: All real \(y\), \((-\infty, \infty)\).
  1. Analysis of Absolute Value (Q5)
  2. Question 5: Absolute Value Function
  • a. \(f(x) = 3 + 2|x|\):
  • \(f(-5) = 3 + 2|-5| = 13\)
  • \(f(0) = 3 + 2|0| = 3\)
  • \(f(-5/2) = 3 + 2|-5/2| = 3+5 = 8\)
  • \(f(3) = 3 + 2|3| = 9\)
  • b. Domain: \((-\infty, \infty)\). Range: Since \(|x| \ge 0\), \(f(x) \ge 3\), so \([3, \infty)\).
  • c. Graph: A V-shaped graph with vertex at \((0, 3)\), passing through \((-1, 5)\) and \((1, 5)\).
  1. Analysis of Signum Function (Q6)
  2. Question 6: Signum Function
  • Definition: \(\operatorname{sgn}(x) = 1\) if \(x>0\), \(0\) if \(x=0\), \(-1\) if \(x<0\).
  • Function: \(f(x) = 2 - 3\operatorname{sgn}(x)\).
  • Example evaluations:
  • For \(x = 5\): \(f(5) = 2 - 3(1) = -1\).
  • For \(x = 0\): \(f(0) = 2 - 3(0) = 2\).
  • For \(x = -5\): \(f(-5) = 2 - 3(-1) = 5\).

Examples

In engineering, understanding relations and their inverses is crucial for control systems. For example, if a sensor relates a physical input (like temperature) to an electrical voltage output, the inverse function allows us to calculate the temperature from a measured voltage. Similarly, absolute value functions and power functions model physical constraints and energy relationships, such as how distance relates to time in free fall or how power varies with electrical resistance. These concepts form the bedrock for modeling real-world phenomena accurately.

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