Lecture 9: Basic Mathematics PDF

# MATH107 BASIC MATHEMATICS ## Last week we finish discussing ### INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences ### Chapter 3 Lines, Parabolas, and Systems ## Chapter 2: Functions and Graphs ### 2.1 Functions - A function assigns each input number t...

# MATH107 BASIC MATHEMATICS ## Last week we finish discussing ### INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences ### Chapter 3 Lines, Parabolas, and Systems ## Chapter 2: Functions and Graphs ### 2.1 Functions - A function assigns each input number to one output number. - The set of all input numbers is the domain of the function. - The set of all output numbers is the range. ### Equality of Functions - Two functions f and g are equal (f=g): 1. Domain of f = domain of g; 2. f(x) = g(x). #### Example 1 - Determining Equality of Functions Determine which of the following functions are equal. - a. $f(x) = \frac{(x + 2)(x - 1)}{(x - 1)}$ - b. $g(x) = x + 2$ - c. $h(x) = \begin{cases} x + 2\text{ if } x\neq1 \\ 0\text{ if } x = 1 \end{cases}$ - d. $k(x) = \begin{cases} x + 2\text{ if } x\neq1 \\ 3\text{ if } x = 1 \end{cases}$ ##### Solution: When x=1, - f(1) ≠ g(1) - f(1) ≠ h(1) - f(1) ≠ k(1) By definition, g(x)=h(x)=k(x) for all x≠1. Since g(1)=3, h(1)=0 and k(1)=3, we conclude that - g=k - g≠h - h≠k #### Example 3 - Finding Domain and Function Values Let $g(x) = 3x^2 - x + 5$. Any real number can be used for $x$, so the domain of $g$ is all real numbers. - **a. Find g(z).** Solution: $g(z) = 3z^2 - z + 5$ - **b. Find g(r^2).** Solution: $g(r^2) = 3(r^2)^2 - r^2 + 5 = 3r^4-r^2+5$ - **c. Find g(x + h).** Solution: $g(x+h) = 3(x+h)^2-(x+h)+5 = 3x^2+6hx+3h^2-x-h+5$ # Chapter 4: MATHEMATICAL FUNCTIONS ## 4.1. FUNCTIONS. ### DEFINITION: (FUNCTIONS) A function is a mathematical rule that assigns to each input value one and only one output value. ### DEFINITION: (DOMAIN / RANGE) - A domain of a function is the set consisting of all possible input values. - The range of a function is the set of all possible output values. #### Example of input, function and output:  #### Example: Consider the equation $y=x^2-2x+1$. * Values of $x$ are input. All $x$ values are domain. * Values of $y$ are output. All possible $y$ values are range. If $x=1$, then $y=1^2-2.1+1=0.$ #### Function diagrams:  #### Example: Let $y=2x^2+5x+6$. Find $f(2)$. $x=2 \implies y=f(2)=2.(2)^2 +5.2+6 = 2.4+10+6 = 24$ $f(2)=24$ #### Example: $t=u(v)=2v^2-5v$. Determine $u(-5)$. $t=u(-5)=2-(-5)^2 - 5.(-5) = 50+25=75$. #### Example: Given: $y=f(x)=x^2-2x+3$ * Domain, $D = \mathbb{R} = \{x|x\in\mathbb{R}\}$ #### Example: Given: $y = f(x) = \frac{1}{x^2-4}$ $x^2-4\neq0$ because $\frac{1}{0}$ is undefined. $x\neq2$ $D=\mathbb{R}\setminus\{-2, 2\}=\mathbb{R}-\{-2, 2\}$ #### Example: $y=f(x)=\sqrt{x-5}$ We know that $\sqrt{g(x)}$ is defined if $g(x)\geq0$ So, $x-5\geq0 \implies x\geq5$ $D = [5, \infty ) = \{x|x\geq5\}$ #### Example: $y = f(x) = \sqrt{25-x^2}$ $25-x^2 \geq 0 \implies (5-x)(5+x) \geq 0$ $D=[-5, 5]$ #### Example: $y=\sqrt{\frac{x+3}{x^2-1}}$ $x+3\geq0\implies x\geq-3$ $x^2-1=0 \implies x^2=1$ $\implies x=\pm1$ $D=(-3,-1)\cup(-1, 1)\cup(1, \infty)$ #### Example: $y=f(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2-x_1x_2x_3$ Find $f(1, 0, -2)=(1)^2 + (0)^2 + (-2)^2 - 1.0.(-2)=5$. ## Chapter 2: Functions and Graphs ### 2.2 Special Functions - We begin with the constant function. #### Example 1 - Constant Function Let $h(x)=2$. The domain of $h$ is all real numbers. $h(10)=2$. $ h(-387)=2$. $h(x+3)=2$. A function of the form $h(x)=c$, where $c$ is a constant, is a constant function. #### Example 3 - Rational Functions - a. $f(x) = \frac{x^2-6x}{x+5}$ is a rational function, since the numerator and denominator are both polynomials. - b. $g(x)=2x+3$ is a rational function, since $2x+3=\frac{2x+3}{1}$. #### Example 5 - Absolute-Value Function Absolute-value function is defined as $|x|$, e.g. $|x| = \begin{cases} x\text{ if } x\leq0\\ -x\text{ if } x<0 \end{cases} $ ## Chapter 2: Functions and Graphs ### 2.3 Combinations of Functions - We define the operations of functions as: * $(f+g)(x)=f(x)+g(x)$ * $(f-g)(x)=f(x)-g(x)$ * $(fg)(x)=f(x).g(x)$ * $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$ for $g(x)\neq0$ #### Example 1 - Combining Functions If $f(x)=3x-1$ and $g(x)=x^2+3x$, find: - a. $(f+g)(x)$ - b. $(f-g)(x)$ - c. $(fg)(x)$ - d. $\left(\frac{f}{g}\right)(x)$ - e. $\left(\frac{1}{2}f\right)(x)$ ##### Solution: - a. $(f+g)(x)=f(x)+g(x)=(3x-1)+(x^2+3x)=x^2+6x-1$ - b. $(f-g)(x)=f(x)-g(x)=(3x-1)-(x^2+3x)=-1-x^2$ - c. $(fg)(x)=f(x)g(x)=(3x-1)(x^2+3x)=3x^3+8x^2-3x$ - d. $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{3x-1}{x^2+3x}$ - e. $\left(\frac{1}{2}f\right)(x)=\frac{1}{2}(f(x))=\frac{1}{2}(3x-1)=\frac{3x-1}{2}$ ### Composition - Composite of f with g is defined by $(f\circ g)(x)=f(g(x))$ #### Example 3 - Composition If $F(p)=p^2+4p-3$, $ G(p) = 2p+1$ and $H(p) = |p|$, find - a. $F(G(p))$ - b. $F(G(H(p)))$ - c. $G(F(1))$ ##### Solution: - a. $F(G(p)) = F(2p+1)=(2p+1)^2+4(2p+1)-3=4p^2+12p+2 = (FG)(p)$ - b. $F(G(H(p)))=(F(GH))(p)=((FG)H)(p) = (FG)(H(p))=(FG)(|p|)=4|p|^2 +12|p| + 2=4p^2+12|p|=2$. - c. $G(F(1)) = G(1^2+4-1-3) = G(2) = 2.2+1 = 5$ ## Chapter 2: Functions and Graphs ### 2.4 Inverse Functions - An inverse function is defined as $f^{-1}(f(x)) = x = f(f^{-1}(x))$ #### Example 1 - Inverses of Linear Functions Show that a linear function is one-to-one. Find the inverse of $f(x)=ax+b$ and show that is also linear. ##### Solution: Assume that $f(u)=f(v)$, thus $au+b=av+b$. We can prove the relationship $(f \circ g)(x) = f(g(x))=a(x-b)+b=x$ $(g\circ f)(x) = g(f(x)) = \frac{(ax+b)-b}{a}=\frac{ax}{a}=x$ #### Example 3 - Inverses Used to Solve Equations Many equations take the form $f(x) = 0$, where $f$ a function. If $f$ is a one-to-one function, then the equation has $x=f^{-1}(0)$ as its unique solution. ##### Solution: Applying $f^{-1}$ to both sides gives $f^{-1}(f(x)) = f^{-1}(0)$. Since $f(f^{-1}(0))=0$, $f^{-1}(0)$ is a solution. #### Example 5 - Finding the Inverse of a Function To find the inverse of a one-to-one function $f$, solve the equation $y=f(x)$ for $x$ in terms of $y$, obtaining $x=g(y)$. Then $f^{-1}(x)=g(x)$. To illustrate, find $f^{-1}(x)$ if $f(x)=(x-1)^2$, for $x\geq 1$. ##### Solution: Let $y=(x-1)^2$, for $x\geq1$. Then $x-1=\sqrt{y}$ and hence $x=\sqrt{y}+1$. It follows that $f^{-1}(x)=\sqrt{x}+1$. #### Example: If $f(x)=3x-5$, $g(x)=x^2-2x+1$, $h(x) = x^3$ and $j(x) = \frac{1}{2x^4}$. Find: - a. $f(x)+g(x)$ - b. $h(x)-j(x)$ - c. $f(x).h(x)$ - d. $\frac{h(x)}{j{x}}$ ##### Solution: - a. $p(x)=f(x)+g(x)=3x-5+x^2-2x+1=x^2+x-4$ - b. $q(x)=h(x)-j(x)=x^3-\frac{1}{2x^4}$ - c. $s(x)=f(x).h(x)=(3x-5)x^3=3x^4-5x^3$ - d. $t(x)=\frac{h(x)}{f(x)}=\frac{x^3}{\frac{1}{2x^4}}=x^3.2x^4=2x^7$ #### Example: $y=g(x)=2x+50$. $x=h(p)=(150-2,5p$. $y=f(p)=g(h(p))$. $\implies y=f(p)=g(150-2,5p)$. $\implies y=2.(150-2,5p)+50$ $\implies y=300-5p+50$ $\implies y=350-5p$ #### Example: If: - $y=g(u)=u^2-2u+10$ - $u=h(x)=x+1$ Then, the composite function: $y = f(x) = g(h(x)) = ?$ ##### Solution: $y=f(x)=g(x+1)=(x+1)^2-2(x+1)+10=x^2+2x+1-2x-2+10=x^2+9$ #### Example: If $y=g(u)=2u^3$ and $u=h(x)=x^2-2x+5$ determine: - a. $g(h(x))$ - b. $g(h(2))$ - c. $g(h(-3))$ ##### Solution: - a. $g(x^2-2x+5)=2(x^2-2x+5)^3$ - b. $g(h(2))=2.(2^2-2.2+5)^3=2.(4-4 +5)^3=2.5^3=250$ - c. $g(h(-3))=2((-3)^2-2.(-3)+5)^3=2.(9+6+5)^3=2.20^3=16000$