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Section 1.3 Functions

Subsection 1. New vocabulary

Subsubsection Definitions

Write a definition or description for each term. You can find answers in Section 1.2 of your textbook.

  1. Function
  2. Input variable
  3. Output variable
  4. Function value
  5. Function notation

Subsubsection Exercise

Identify each term above, or give an example, for this situation: At time \(t\) seconds, the height of a basketball above the ground, \(h\text{,}\) in feet, is given by

\begin{equation*} ~~h=-16t^2+20t+5\text{.} \end{equation*}
Answer
  1. \(h\) is a function of \(t\text{.}\)
  2. The input variable is \(t\text{.}\)
  3. The output variable is \(h\text{.}\)
  4. The function value for \(t=1\) is \(h=9\text{.}\)
  5. \(\displaystyle h=f(t)\)

Subsection 2. Solve linear equations and inequalities with parentheses

Strategy for solving linear equations.
  1. Simplify each side of the equation: apply the distributive law, combine like terms.
  2. Use addition and subtraction to get all the variable terms on one side of the equation, and all constsnt terms on the other side.
  3. Divide both sides by the coefficient of the variable.

Subsubsection Examples

Example 1.47.

Solve \(~~3(2a-4) \ge 4-(1-3a)\)

Solution

First, we remove parentheses by applying the distributive law. Then we can combine like terms on each side of the equation.

Note that the minus sign in front of the parentheses on the right side of the equation applies to both terms inside the parentheses.

\begin{align*} 3(2a-4) \amp \ge 4-(1-3a) \amp \amp \blert{\text{Apply the distributive law.}}\\ 6a-12 \amp \ge 4\alert{-}1\alert{+}3a \amp \amp \blert{\text{Simplify the right side.}}\\ 6a-12 \amp \ge 3+3a \amp \amp \blert{\text{Subtract}~ 3a~ \text{from both sides.}}\\ 3a-12 \amp \ge 3 \amp \amp \blert{\text{Add 12 to both sides.}}\\ 3a \amp \ge 15 \amp \amp \blert{\text{Divide both sides by 3.}}\\ a \amp \ge 5 \end{align*}
Example 1.48.

Solve the inequality \(~~25-6x \gt 3x-2(4-x)\)

Solution

We begin by the same way we solve an equation. For this example, we start by removing the parentheses.

\begin{align*} 25-6x \amp \gt 3x-2(4-x) \amp \amp \blert{\text{Apply the distributive law.}}\\ 25-6x \amp \gt 3x-8+2x \amp \amp \blert{\text{Combine like terms.}}\\ 25-6x \amp \gt 5x-8 \amp \amp \blert{\text{Subtract}~ 5x~ \text{from both sides.}}\\ 25-11x \amp \gt -8 \amp \amp \blert{\text{Subtract 25 from both sides.}}\\ -11x \amp \gt -33 \amp \amp \blert{\text{Divide both sides by -11.}}\\ x \amp \alert{\lt} 3 \amp \amp \blert{\text{Don't forget to reverse the inequality symbol.}} \end{align*}

Recall that if we multiply or divide both sides of an inequality by a negative number, we must reverse the direction of the inequality symbol.

Subsubsection Exercises

Solve the inequality \(~~-4(x+2)+3(x-2) \ge -2\)

Answer
\(x \le -12\)

Solve the equation \(~~4(2-3w)=9-3(2w-1)\)

Answer
\(\dfrac{-1}{2}\)

Solve the inequality \(~~2(3h-6) \lt 5-(h-4)\)

Answer
\(h \lt 3\)

Solve the equation \(~~0.25(x+3)-0.45(x-3)=0.30\)

Answer
\(9\)

Subsection 3. Solve non-linear equations

To solve simple non-linear equations, we "undo" the operation performed on the variable.

Subsubsection Examples

Example 1.53.

Solve the equation \(~~5 \sqrt{t} = 83\)

Solution

To "undo" a square root, we square both sides of the equation. First, we isolate the square root.

\begin{align*} \dfrac{5 \sqrt{t}}{\alert{5}} \amp = \dfrac{83}{\alert{5}} \amp \amp \blert{\text{Divide both sides by 5.}}\\ (\sqrt{t})^2 \amp =(16.6)^2 \amp \amp \blert{\text{Square both sides.}}\\ t \amp =275.56 \end{align*}
Example 1.54.

Solve the equation \(~~\dfrac{15}{y}=45\)

Solution

If the variable is in the denominator of a fraction, we must first clear the fraction.

\begin{align*} \alert{y}(\dfrac{15}{y}) \amp = 45 \cdot \alert{y} \amp \amp \blert{\text{Multiply both sides by} ~y.}\\ 15 \amp = 45y \amp \amp \blert{\text{Divide both sides by 45.}}\\ y \amp =\dfrac{15}{45}=\dfrac{1}{3} \end{align*}

Subsubsection Exercises

Solve the equation \(~~\dfrac{4.8}{w}=3\)

Answer
\(1.6\)

Solve the equation \(~~18=36\sqrt{q}\)

Answer
\(\dfrac{1}{4}\)

Subsection 4. Working with exponents

Recall the laws of exponents:

Laws of Exponents.

  1. \(\displaystyle a^m\cdot a^n = a^{m+n}~~~~\blert{\text{Product of Powers}}\)

    1. \(\displaystyle \dfrac{a^m}{a^n}=a^{m-n} \hphantom{blank}m\gt n~~~~\blert{\text{Quotient of Powers}}\)

    2. \(\displaystyle \displaystyle{\frac{a^m}{a^n}=\frac{1}{a^{n-m}} \hphantom{blank}m\lt n}\)

  3. \(\displaystyle \left(a^m\right)^n=a^{m+n}~~~~\blert{\text{Power of a Power}}\)

  4. \(\displaystyle (ab)^n=a^n b^n~~~~\blert{\text{Power of a Product}}\)

  5. \(\displaystyle \displaystyle{\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} }~~~~\blert{\text{Power of a Quotient}}\)

Subsubsection Examples

Example 1.57.

Here are some examples of the correct use of the laws of exponents.

  1. \(\displaystyle x^3 \cdot x^5 = x^8~~~~ \blert{\text{We add the exponents when multiplying.}}\)
  2. \(\displaystyle (x^3)^5 = x^{15}~~~~ \blert{\text{To raise a power to a power, we multiply exponents.}}\)
  3. \(\displaystyle \dfrac{x^3}{x^5} = \dfrac{1}{x^2}~~~~\blert{\text{To divide, we subtract exponents.}}\)
  4. \(\displaystyle (xy)^3 = x^3y^3~~~~\blert{\text{The power of a product is the product of the powers.}}\)
Example 1.58.

Here are some \(\alert{\text{MISTAKES}}\) to avoid.

  1. \(\displaystyle x^3 \cdot x^5 \not= x^{15}~~~~\blert{\text{We should add the exponents.}}\)
  2. \(\displaystyle (2x)^3 \not= 2x^3~~~~\blert{\text{2 is also cubed.}}\)
  3. \(\displaystyle (x+2)^3 \not= x^3+8~~~~\blert{\text{The product rule does not apply to sums.}}\)
  4. \(\displaystyle 2 \cdot 5^3 \not= 10^3~~~~\blert{\text{We compute powers before products.}}\)

Subsubsection Exercises

Simplify each expression.

  1. \(\displaystyle x^2(x^2)^3\)
  2. \(\displaystyle (2t^2)^4\)
  3. \(\displaystyle \dfrac{5^6}{5^2}\)
  4. \(\displaystyle (-h)^3-h^3\)
Answer
  1. \(\displaystyle x^2(x^6) = x^8\)
  2. \(\displaystyle 2^4(t^2)^4 = 16t^8\)
  3. \(\displaystyle 5^4~~~~\blert{\text{Subtract exponents; keep the same base.}}\)
  4. \(\displaystyle -h^3-h^3 = -2h^3\)

Each "simplification" is INCORRECT. Write a correct version.

  1. \(\displaystyle (3+b^3)^2 \rightarrow 9+b^6\)
  2. \(\displaystyle 5a^3+3a^2 \rightarrow 8a^5\)
  3. \(\displaystyle (4x^4)^2 \rightarrow 16x^{16}\)
  4. \(\displaystyle \dfrac{w^3}{w^9} \rightarrow -w^3\)
Answer
  1. \(\displaystyle 9+6b^3+b^6\)
  2. cannot be simplified
  3. \(\displaystyle 16x^8\)
  4. \(\displaystyle \dfrac{1}{w^6}\)