MA 161 - Lesson 2 (1.3)

Original Notes

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MA 161 - Lesson 2 (1.3)

Today:

  • Inverse of Function
  • One to One Function
  • Horizontal Line Test
  • Exponential and Logarithmic Functions
  • Properties of Logarithmic functions

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Warmup:

\[ \overset{\text{output}}{y} = f(\overset{\text{input}}{x}) = x^3 \]

find the value of input when \(\text{output} = 8\)

find \( x \), given \( y = 8 \)

\[ 8 = y = x^3 \rightsquigarrow x = 8^{1/3} = 2 \] \[ y = x^3 \implies x = y^{1/3} \]

Let \( g(y) = y^{1/3} \rightsquigarrow \) called inverse of the function \( f(x) = x^3 \)

Denoted by

\[ f^{-1}(x) = x^{1/3} \]

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eg: \( f(x) = x^2 \)

find input when \(\text{output} = 4\)

two possible inputs ie \( x = 2 \) or \( x = -2 \)

Inverse does not exist

Question: what guarantees existence of an inverse.

\( f(x) \) has to be \( 1\text{-}1 \)

ie each output has exactly one input

then \( f^{-1} \) exist

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Horizontal Line Test \(\rightsquigarrow\) Check whether given graph is 1-1 or Not

eg: \(f(x) = x^3\)

Cartesian coordinate graph of the cubic function f(x) = x^3, shown as a blue curve passing through the origin. Three horizontal red lines are drawn across the plot, each intersecting the curve at exactly one point, illustrating that the function passes the horizontal line test.
Visual Description: Cartesian coordinate graph of the cubic function f(x) = x^3, shown as a blue curve passing through the origin. Three horizontal red lines are drawn across the plot, each intersecting the curve at exactly one point, illustrating that the function passes the horizontal line test.

All horizontal lines intersect at exactly one point

\(\Rightarrow f\) is 1-1, invertible

eg: \(f(x) = x^2\)

Cartesian coordinate graph of the quadratic function f(x) = x^2, shown as a blue upward-opening parabola with its vertex at the origin. Multiple horizontal red lines are drawn across the plot, with the middle line intersecting the parabola at two distinct points, illustrating that the function fails the horizontal line test.
Visual Description: Cartesian coordinate graph of the quadratic function f(x) = x^2, shown as a blue upward-opening parabola with its vertex at the origin. Multiple horizontal red lines are drawn across the plot, with the middle line intersecting the parabola at two distinct points, illustrating that the function fails the horizontal line test.

There is a horizontal line that intersects at more than one point

\(\Rightarrow f\) is not 1-1, not invertible.

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Restricting Domain to Make a Function 1-1

Consider the function \( f(x) = x^2 \):

Cartesian coordinate graph of the parabola f(x) = x^2 with its vertex at the origin. The left half of the parabola for negative x is drawn in green, and the right half for positive x is drawn in blue. Multiple horizontal red line segments cross the curve, illustrating the horizontal line test showing that the unrestricted function is not one-to-one, while restricting the domain to either positive or negative x makes it one-to-one.
Visual Description: Cartesian coordinate graph of the parabola f(x) = x^2 with its vertex at the origin. The left half of the parabola for negative x is drawn in green, and the right half for positive x is drawn in blue. Multiple horizontal red line segments cross the curve, illustrating the horizontal line test showing that the unrestricted function is not one-to-one, while restricting the domain to either positive or negative x makes it one-to-one.

  1. Restrict domain to \( [0, \infty) \):

    \( y = x^2 \) is one-to-one (1-1) and invertible.

    \( \Rightarrow x = \sqrt{y} \)

    \( f^{-1}(y) = \sqrt{y} \)

    • Domain of \( f^{-1} \): \( [0, \infty) \)
    • Range of \( f^{-1} \): \( [0, \infty) \)

  2. Restrict domain to \( (-\infty, 0] \):

    \( y = x^2 \) is one-to-one (1-1) and invertible.

    \( x = -\sqrt{y} \) is the input, given an output.

    \( f^{-1}(y) = -\sqrt{y} \)

    • Domain of \( f^{-1} \): \( [0, \infty) \)
    • Range of \( f^{-1} \): \( (-\infty, 0] \)

Note: The domain and range of \( f \) and \( f^{-1} \) are switched.

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Example: \( f(x) = |x - 5| \)

\( D: (-\infty, \infty) \rightsquigarrow f \) is not 1-1 \( \Rightarrow \) not invertible

Graph of the absolute value function f(x) = |x - 5| plotted on a Cartesian plane. The V-shaped graph has its vertex on the x-axis at x = 5. The left ray (green) decreases to (5, 0), and the right ray (blue) increases from (5, 0). A horizontal dashed purple line intersects both branches at height y, illustrating that the function fails the horizontal line test over the full domain (-infinity, infinity). Multiple red horizontal bars along the right branch highlight the restricted domain [5, infinity) where the function is one-to-one, with a vertical dashed purple line dropping down to the x-axis.
Visual Description: Graph of the absolute value function f(x) = |x - 5| plotted on a Cartesian plane. The V-shaped graph has its vertex on the x-axis at x = 5. The left ray (green) decreases to (5, 0), and the right ray (blue) increases from (5, 0). A horizontal dashed purple line intersects both branches at height y, illustrating that the function fails the horizontal line test over the full domain (-infinity, infinity). Multiple red horizontal bars along the right branch highlight the restricted domain [5, infinity) where the function is one-to-one, with a vertical dashed purple line dropping down to the x-axis.
1) On \( D: [5, \infty) \), \( f \) is 1-1, invertible

Given \( y \), what is \( x \)?

\[ y = x - 5 \Rightarrow x = y + 5 \]

\[ \vec{f}(y) = y + 5 \rightsquigarrow D: [0, \infty) \]

\[ R: [5, \infty) \]

2) On \( D: (-\infty, 5] \), \( f \) is 1-1, invertible

Given \( y \), what is \( x \)?

\[ x < 5 \rightsquigarrow x - 5 < 0 \rightsquigarrow |x - 5| = -(x - 5) > 0 \]

\[ y = -(x - 5) = -x + 5 \rightsquigarrow x = 5 - y \]

\[ \vec{f}(y) = 5 - y \rightsquigarrow D: [0, \infty), R: (-\infty, 5] \]

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\[ |5| = 5 \]

\[ |-5| = 5 = -(-5) \]

\[ |x| = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases} \]

\[ |x-5| = \begin{cases} x-5 & x-5 \ge 0 \\ -(x-5) & x-5 < 0 \end{cases} \]

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eg: \( f(x) = \frac{1}{x-3} \), is \( f(x) \) 1-1, if so find \( f^{-1}(x) \).

Graph of the rational function f(x) = 1/(x-3) on Cartesian coordinate axes. A vertical red dashed line indicates the vertical asymptote at x = 3. The curve consists of two branches: a blue branch in the upper region to the right of x = 3 decreasing towards the horizontal asymptote y = 0, and a green branch in the lower region to the left of x = 3 decreasing towards negative infinity as it approaches the asymptote. Several horizontal red lines are drawn across the graph, demonstrating the horizontal line test where each horizontal line intersects the curve at most once, proving the function is one-to-one.
Visual Description: Graph of the rational function f(x) = 1/(x-3) on Cartesian coordinate axes. A vertical red dashed line indicates the vertical asymptote at x = 3. The curve consists of two branches: a blue branch in the upper region to the right of x = 3 decreasing towards the horizontal asymptote y = 0, and a green branch in the lower region to the left of x = 3 decreasing towards negative infinity as it approaches the asymptote. Several horizontal red lines are drawn across the graph, demonstrating the horizontal line test where each horizontal line intersects the curve at most once, proving the function is one-to-one.

\( f(x) \):
D: \( (-\infty, 3) \cup (3, \infty) \)
R: \( (-\infty, 0) \cup (0, \infty) \)

\( f \) is 1-1, so invertible

Given \( y \), find \( x \)

\[ y = \frac{1}{x-3}, \quad x \neq 3, \; y \neq 0 \]

\[ x - 3 = \frac{1}{y} \implies x = \frac{1}{y} + 3 \]

\[ f^{-1}(y) = \frac{1}{y} + 3 \]

D: \( (-\infty, 0) \cup (0, \infty) \)
R: \( (-\infty, 3) \cup (3, \infty) \)

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Exponential and Logarithmic Functions

\[ f(x) = b^x \]

\[ b > 0, b \neq 1 \]

Graph of an exponential function f(x) = b^x on the Cartesian plane. The blue curve is strictly increasing, passing through the y-intercept at (0, 1), and asymptotically approaching zero along the negative x-axis. Several horizontal red lines cross the curve, demonstrating that it intersects every horizontal line at most once, verifying that the function is one-to-one via the horizontal line test.
Visual Description: Graph of an exponential function f(x) = b^x on the Cartesian plane. The blue curve is strictly increasing, passing through the y-intercept at (0, 1), and asymptotically approaching zero along the negative x-axis. Several horizontal red lines cross the curve, demonstrating that it intersects every horizontal line at most once, verifying that the function is one-to-one via the horizontal line test.

\( f \) is 1-1
\( \Rightarrow \) invertible

\( D: (-\infty, \infty) \)
\( R: (0, \infty) \)

\[ y = b^x \leadsto x = \log_b y \]

Graph of a logarithmic function f^{-1}(x) = \log_b(x) on the Cartesian plane. The green curve is strictly increasing, passing through the x-intercept at (1, 0), with a vertical asymptote along the negative y-axis as x approaches zero from the right, extending towards positive infinity as x increases.
Visual Description: Graph of a logarithmic function f^{-1}(x) = \log_b(x) on the Cartesian plane. The green curve is strictly increasing, passing through the x-intercept at (1, 0), with a vertical asymptote along the negative y-axis as x approaches zero from the right, extending towards positive infinity as x increases.

\( D: (0, \infty) \)
\( R: (-\infty, \infty) \)

\[ f^{-1}(x) = \log_b x \]

is the inverse of \( b^x \)
called logarithm to base \( b \).

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Properties of Logarithms

  1. \[ \log_b (xy) = \log_b x + \log_b y \]
  2. \[ \log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y \]
  3. \[ \log_b x^r = r \log_b x \]
  4. \[ \log_b b = 1, \quad b^1 = b \]
  5. \[ \log_b 1 = 0, \quad b^0 = 1 \]
  6. \[ \log_b b^x = x = b^{\log_b x} \]

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Eg: Find \(x\) such that \(3^{7x-2} = 18\)

\[ \log_b b^r = r \log_b b = r \]

take \(\log\) to base \(3\) on both sides

\[ \log_3 3^{7x-2} = \log_3 18 \] \[ (7x-2)\underbrace{\log_3 3}_{1} = \log_3 18 \] \[ 7x - 2 = \log_3 18 \] \[ x = \frac{\log_3 18 + 2}{7} = \frac{\log_3 2 + 4}{7} \] \[ \log_3 18 = \log_3(2 \times 9) = \log_3 2 + \log_3 9 = \log_3 2 + \log_3 3^2 = \log_3 2 + 2 \]

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e.g.: Write \(6^{4x}\) as exponential base \(e\).

Recall!

\[ e^{\log_e \square} = \square \] \[ e^{\log_e 5} = 5 \] \[ e^{\log_e 6^{4x}} = 6^{4x} \] \[ \rightsquigarrow \quad 6^{4x} = e^{4x \log_e 6} \]