Section 1.6: Inverse Functions, Logarithms, and Inverse Trigonometry

Course Announcements

  • Knowledge Knockouts are due Saturday, 9/5.
  • Homework 1 (Section 1.6) is due Wednesday, 8/26 by 11:59 pm.
  • Supplemental Instruction (SI) starts next week.

Algebra Review: Functions, Domain, and Range

  • Definition of a Function: A function f(x)f(x) is a rule that assigns to each element xx in a set DD exactly one element, called f(x)f(x), in a set RR.
    • The set DD is called the domain of the function.
    • The set RR is called the range of the function.
  • Common Algebraic Functions and Their Domains:
    • Polynomial Functions:
      • Example: f(x)=7x4+4x2−x+3f(x) = 7x^4 + 4x^2 - x + 3
      • Domain: (−∞,∞)(-\infty, \infty)
    • Radical Functions:
      • Even roots require non-negative radicands (radicand≥0\text{radicand} \ge 0).
      • Example 1 (Square Root): f(x)=3x−6f(x) = \sqrt{3x - 6}
        • Domain inequality: 3x−6≥0⇒3x≥6⇒x≥23x - 6 \ge 0 \Rightarrow 3x \ge 6 \Rightarrow x \ge 2
        • Domain in interval notation: [2,∞)[2, \infty)
      • Example 2 (Cube Root): f(x)=9−7x23f(x) = \sqrt[3]{9 - 7x^2}
        • Domain: (−∞,∞)(-\infty, \infty) because odd roots accept all real numbers.
    • Rational Functions:
      • Rational functions require a non-zero denominator (denominator≠0\text{denominator} \neq 0).
      • Example: f(x)=x+1x−5f(x) = \frac{x + 1}{x - 5}
        • Domain condition: x−5≠0⇒x≠5x - 5 \neq 0 \Rightarrow x \neq 5
        • Domain in interval notation: (−∞,5)∪(5,∞)(-\infty, 5) \cup (5, \infty)

One-to-One Functions and the Horizontal Line Test

  • Definition of One-to-One: A function f(x)f(x) is one-to-one on a domain DD if f(x1)≠f(x2)f(x_1) \neq f(x_2) whenever x1≠x2x_1 \neq x_2 in DD. Equivalently, f(x1)=f(x2)f(x_1) = f(x_2) implies x1=x2x_1 = x_2.
  • The Horizontal Line Test: A function y=f(x)y = f(x) is one-to-one if and only if its graph intersects each horizontal line at most once.
  • Examples of One-to-One and Non-One-to-One Functions:
    • y=xy = \sqrt{x} is one-to-one because its graph intersects every horizontal line at most once.
    • y=x2y = x^2 is not one-to-one because a single horizontal line (such as y=1y = 1) intersects the graph at multiple points (x=−1x = -1 and x=1x = 1).
    • y=sin⁡(x)y = \sin(x) is not one-to-one because it is periodic and repeats output values (e.g., at x=5π36x = \frac{5\pi}{36} and other values), causing horizontal lines to intersect the graph repeatedly.

Inverse Functions: Notation, Properties, and Graphs

  • Inverse vs. Reciprocal Notation:
    • f−1(x)f^{-1}(x) denotes the inverse function.
    • 1f(x)\frac{1}{f(x)} denotes the reciprocal function.
    • Caution: f−1(x)≠1f(x)f^{-1}(x) \neq \frac{1}{f(x)}. The exponent −1-1 in function notation represents the inverse transformation, not exponentiation.
  • Fundamental Inverse Concept:
    • An inverse function interchanges the inputs (xx) and outputs (yy).
    • If f(a)=bf(a) = b, then f^{-1}(b) = a$.\n* **Example 1:**\n * a. If fisaone−to−onefunctionandis a one-to-one function andf(1) = 5,whatis, what isf^{-1}(5)?\n * Solution: f^{-1}(5) = 1\n * b. If fisaone−to−onefunctionandis a one-to-one function andf^{-1}(8) = -10,whatis, what isf(-10)?\n * Solution: f(-10) = 8\n* **Domain and Range Relationship:**\n * \text{Domain of } f^{-1} = \text{Range of } f\n * \text{Range of } f^{-1} = \text{Domain of } f\n* **Graphical Properties of Inverses:**\n * To evaluate f(x)atinputat inputx,startat, start atxonthehorizontalaxis,moveverticallytothecurve,andmovehorizontallytotheon the horizontal axis, move vertically to the curve, and move horizontally to they-axis.\n * To evaluate f^{-1}(y)atvalueat valuey,startat, start atyontheverticalaxis,movehorizontallytothecurve,andmoveverticallydowntotheon the vertical axis, move horizontally to the curve, and move vertically down to thex-axis.\n * The graph of y = f^{-1}(x)isobtainedbyreflectingthegraphofis obtained by reflecting the graph ofy = f(x)acrossthelineacross the liney = x$.
    • Any point (a,b)(a, b) on the graph of y=f(x)y = f(x) maps to the point (b,a)(b, a) on the graph of y=f−1(x)y = f^{-1}(x).
  • Example 2:
    • Given the one-to-one function f(x)=x3f(x) = x^3, sketch its inverse.
    • The inverse function is f−1(x)=x3f^{-1}(x) = \sqrt[3]{x}.
    • Reflecting the cubic curve y=x3y = x^3 across the line y=xy = x yields the graph of the cube root function y=x3y = \sqrt[3]{x}.

Finding a Function's Inverse Algebraically

  • Four-Step Procedure to Find f−1(x)f^{-1}(x):
    1. Replace f(x)f(x) with y$.\n 2. Interchange xandandy$.
    2. Solve the resulting equation for y$.\n 4. Replace ywithwithf^{-1}(x).\n* **Example 3:**\n * Find the inverse of the one-to-one function f(x) = 5x - 7$.
    • Step 1: y=5x−7y = 5x - 7
    • Step 2: x=5y−7x = 5y - 7
    • Step 3: Solve for yy
      • x+7=5yx + 7 = 5y
      • y=x+75y = \frac{x + 7}{5}
    • Step 4: f−1(x)=x+75f^{-1}(x) = \frac{x + 7}{5} (or f−1(x)=15x+75f^{-1}(x) = \frac{1}{5}x + \frac{7}{5})
  • Example 4:
    • Find the inverse of the one-to-one function f(x) = 8 - \frac{1}{3}\sqrt[5]{x^3 + 7}$.\n * Step 1: y = 8 - \frac{1}{3}\sqrt[5]{x^3 + 7}\n * Step 2: Interchange xandandy\n * x = 8 - \frac{1}{3}\sqrt[5]{y^3 + 7}\n * Step 3: Solve for y\n * x - 8 = -\frac{1}{3}\sqrt[5]{y^3 + 7}\n * -3(x - 8) = \sqrt[5]{y^3 + 7}\n * -3x + 24 = \sqrt[5]{y^3 + 7}\n * (-3x + 24)^5 = y^3 + 7\n * y^3 = (-3x + 24)^5 - 7\n * y = \sqrt[3]{(-3x + 24)^5 - 7}\n * Step 4: f^{-1}(x) = \sqrt[3]{(-3x + 24)^5 - 7}\n\n# Exponential and Logarithmic Functions\n\n* **General Exponential Function:**\n * Form: f(x) = a^xwherewherea > 0\n * Domain: (-\infty, \infty)\n * Range: (0, \infty)\n* **General Logarithmic Function:**\n * Form: f(x) = \log_a(x)wherewherea > 0andanda \neq 1\n * Defines the inverse function of f(x) = a^x\n * Domain: (0, \infty)\n * Range: (-\infty, \infty)\n* **Natural Exponential Function:**\n * Form: f(x) = e^x\n * It is a one-to-one function.\n * Domain: (-\infty, \infty)\n * Range: (0, \infty)\n * Passes through the point (0, 1).\n* **Natural Logarithmic Function:**\n * Form: f(x) = \ln(x),where, wheree is the understood base.\n * It is the inverse function of f(x) = e^x$.
    • Domain: (0,∞)(0, \infty)
    • Range: (−∞,∞)(-\infty, \infty)
    • Passes through the point (1,0)(1, 0).

Algebraic Properties of the Natural Logarithm

  • Theorem 1: Algebraic Properties of the Natural Logarithm
    • For any positive numbers b>0b > 0 and x>0x > 0, the natural logarithm satisfies the following rules:
    • 1. Product Rule: ln⁡(bx)=ln⁡(b)+ln⁡(x)\ln(bx) = \ln(b) + \ln(x)
    • 2. Quotient Rule: ln⁡(bx)=ln⁡(b)−ln⁡(x)\ln\left(\frac{b}{x}\right) = \ln(b) - \ln(x)
    • 3. Reciprocal Rule: ln⁡(1x)=−ln⁡(x)\ln\left(\frac{1}{x}\right) = -\ln(x) (derived from Rule 2 with b=1b = 1)
    • 4. Power Rule: ln⁡(xr)=rln⁡(x)\ln(x^r) = r\ln(x)
  • Example 5 (Condensing Logarithmic Expressions):
    • Write as a single term: 3ln⁡(3t2−1)−ln⁡(t+1)3\ln(3t^2 - 1) - \ln(t + 1)
    • Apply Power Rule to the coefficient: ln⁡((3t2−1)3)−ln⁡(t+1)\ln((3t^2 - 1)^3) - \ln(t + 1)
    • Apply Quotient Rule to condense: ln⁡((3t2−1)3t+1)\ln\left(\frac{(3t^2 - 1)^3}{t + 1}\right)
  • Example 6 (Expanding Logarithmic Expressions):
    • Expand the expression: ln⁡(xx2+2(3x+1)(x−4)2)\ln\left(\frac{x\sqrt{x^2 + 2}}{(3x + 1)(x - 4)^2}\right)
    • Rewrite radical terms as fractional exponents: ln⁡(x(x2+2)1/2(3x+1)(x−4)2)\ln\left(\frac{x(x^2 + 2)^{1/2}}{(3x + 1)(x - 4)^2}\right)
    • Apply Product and Quotient Rules: ln⁡(x)+ln⁡((x2+2)1/2)−ln⁡(3x+1)−ln⁡((x−4)2)\ln(x) + \ln((x^2 + 2)^{1/2}) - \ln(3x + 1) - \ln((x - 4)^2)
    • Apply Power Rule to bring exponents to the front: ln⁡(x)+12ln⁡(x2+2)−ln⁡(3x+1)−2ln⁡(x−4)\ln(x) + \frac{1}{2}\ln(x^2 + 2) - \ln(3x + 1) - 2\ln(x - 4)

Trigonometric Functions and Their Inverses

  • Standard Trigonometric Functions:
    • The six primary trigonometric functions are sin⁡(x)\sin(x), cos⁡(x)\cos(x), tan⁡(x)\tan(x), csc⁡(x)\csc(x), sec⁡(x)\sec(x), and cot⁡(x)\cot(x).
    • These functions are periodic and repeat their outputs, making them fail the Horizontal Line Test over their full domains.
  • Restricting Domains for Inverses:
    • By restricting the domains of trigonometric functions, they become one-to-one and possess inverse functions.
    • Statement equivalency: sin⁡−1(x)=y  ⟺  sin⁡(y)=x\sin^{-1}(x) = y \iff \sin(y) = x
    • Alternative notation: y=arcsin⁡(x)y = \arcsin(x), which explicitly distinguishes inverse functions from reciprocal functions (csc⁡(x)=1sin⁡(x)\csc(x) = \frac{1}{\sin(x)}).
  • Definitions and Intervals for Inverse Sine and Cosine:
    • y=arcsin⁡(x)y = \arcsin(x) is the unique number in the interval \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] for which sin⁡(y)=x\sin(y) = x.
    • y=arccos⁡(x)y = \arccos(x) is the unique number in the interval [0,π][0, \pi] for which cos⁡(y)=x\cos(y) = x.
  • Example 7 (Evaluating Exact Values):
    • a. Find the exact value of cos⁡−1(−1)\cos^{-1}(-1)
      • Find y∈[0,π]y \in [0, \pi] such that cos⁡(y)=−1\cos(y) = -1
      • Result: π\pi
    • b. Find the exact value of arctan⁡(−1)\arctan(-1)
      • Find y∈(−π2,π2)y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) such that tan⁡(y)=−1\tan(y) = -1
      • Result: −π4-\frac{\pi}{4}
    • c. Find the exact value of arcsin⁡(sin⁡(a))\arcsin(\sin(a))
      • Assuming a∈[−π2,π2]a \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], the inverse and direct operations cancel out.
      • Result: aa