MATH 110 GENERAL EXAMPLES Example: Find the domain of the given graph of the function ๐ฆ 420 5 -2- Example: Find the range of the given graph of the function 1 ๐ฅ Example: If ๐(๐ฅ ) = 2๐ฅ 2 − 4 , compute ๐(๐ + 1) 2 Example: Determine whether the below curves are the graph of a function of x By the vertical line test: A curve in xy-plane is a graph of a function of x if and only if no vertical line intersects the curve more than once. 3 Even and Odd functions: Let ๐(๐ฅ ) be a defined function, then ๏ท ๐ (๐ฅ ) is an even function if ๐(−๐ฅ ) = ๐(๐ฅ ) , ∀ ๐ฅ ∈ ๐ท๐ ๏ท ๐ (๐ฅ ) is an odd function if ๐(−๐ฅ ) = −๐ (๐ฅ ) , ∀ ๐ฅ ∈ ๐ท๐ Useful Rules: ๐๐๐ 1) ๐๐ฃ๐๐ = ๐๐๐ ๐๐ฃ๐๐ 3) ๐๐ฃ๐๐ = ๐๐ฃ๐๐ 5) (−1)0๐๐ ๐๐ข๐๐๐๐ = −1 4 2) ๐๐ฃ๐๐ ๐๐๐ = ๐๐๐ ๐๐๐ 4) ๐๐๐ = ๐๐ฃ๐๐ 6) (−1)๐๐ฃ๐๐ ๐๐ข๐๐๐๐ = 1 Example: Determine whether ๐ (๐ฅ ) is even, odd, or neither even nor. odd functions. 1) ๐(๐ฅ ) = ๐ฅ 5 + 2๐ฅ 2) ๐(๐ฅ ) = 4๐ฅ 2 − ๐ฅ 4 + 4 ๐ฅ 2 +1 3) ๐(๐ฅ ) = ๐ฅ 7 −3๐ฅ 2 ๐ฅ 3 −4๐ฅ 4) ๐(๐ฅ ) = ๐ฅ 5 +2๐ฅ 9 5 Type of functions: Polynomials: General formula: ๐(๐ฅ ) = ๐๐ ๐ฅ ๐ + ๐๐−1 ๐ฅ ๐−1 + โฏ + ๐2 ๐ฅ 2 + ๐1 ๐ฅ + ๐0 Where ๐ is a nonnegative integer ๐0 , ๐1 , ๐2 , … ๐๐ ∈ R (are constants called coefficients) The degree of the polynomial is ๐ Domain of polynomials: is ๐ = (−∞, ∞) Examples of polynomials: 1 ๐(๐ฅ ) = 4๐ฅ 3 − 2๐ฅ 2 + 2 ๐ฅ − √2 5 ๐(๐ฅ ) = √3๐ฅ 2 + 3 ๐ฅ + 1 Examples of not polynomials: ๐ (๐ฅ ) = ๐ฅ −2 + 4๐ฅ − 1 1 2 ๐ (๐ฅ ) = ๐ฅ + 4๐ฅ + √3 6 Example: Specify the degree of the polynomial ๐(๐ฅ ) = 4๐ฅ 2 + 7๐ฅ 5 − 1 Four types of polynomials: 7 1) ๐(๐ฅ ) = ๐0 โน Constant function 2) ๐(๐ฅ ) = ๐1 ๐ฅ + ๐0 โน Linear function 3) ๐(๐ฅ ) = ๐2 ๐ฅ 2 + ๐1 ๐ฅ + ๐0 โน Quadratic function 4) ๐(๐ฅ ) = ๐3 ๐ฅ 3 + ๐2 ๐ฅ 2 + ๐1 ๐ฅ + ๐0 โน Cubic function Example: State the name of polynomials for the following functions and specify its degree. 1 1) ๐(๐ฅ ) = 2 3 2) ๐(๐ฅ ) = 2 ๐ฅ 2 + ๐ฅ + 3 3) ๐(๐ฅ ) = 4๐ฅ 2 + 1 4) ๐(๐ฅ ) = ๐ฅ 3 − 3๐ฅ 2 8 5) ๐(๐ฅ ) = 5๐ฅ − 4 6) ๐(๐ฅ ) = 3๐ฅ − √3๐ฅ 2 9 Power function: General formula: ๐ (๐ฅ ) = ๐ฅ ๐ , ๐คโ๐๐๐ ๐ ∈ ๐ Examples: ๐(๐ฅ ) = ๐ฅ 3 , ๐ (๐ฅ ) = ๐ฅ ๐(๐ฅ ) = ๐ฅ 1.3 1 2 , ๐ (๐ฅ ) = ๐ฅ −2 , ๐(๐ฅ ) = ๐ฅ √3 Root (radical) Functions: General formula: ๐ (๐ฅ ) = ๐√๐(๐ฅ) , ๐คโ๐๐๐ ๐(๐ฅ ) ๐๐ ๐๐๐๐ฆ๐๐๐๐๐๐ and ๐ ≥ 2 10 Example: Determine the root functions from the following functions 3 1) ๐(๐ฅ ) = √2๐ฅ 2 + 4๐ฅ − 1 2) ๐(๐ฅ ) = √๐ฅ −3 + 4๐ฅ 1 4 3) ๐(๐ฅ ) = √๐ฅ 2 + ๐ฅ 5 − 2 ๐ฅ + 3 4 4) ๐(๐ฅ ) = √√2๐ฅ 3 + 3 ๐ฅ 6 1 5) ๐(๐ฅ ) = √4๐ฅ 2 − ๐ฅ 3 1 6) ๐(๐ฅ ) = √6๐ฅ 2 − 4๐ฅ + 3 11 Domain of the Root functions: ๐ ๐(๐ฅ ) = √๐(๐ฅ) = ๐ = (−∞, ∞) ๐๐ ๐ ๐๐ ๐๐ ๐ ๐(๐ฅ ) ≥ 0 ๐๐ ๐ ๐๐ ๐๐๐๐ Example: Find the domain of the below functions: 1) ๐(๐ฅ ) = √๐ฅ − 1 3 2) ๐(๐ฅ ) = √6๐ฅ 2 − 1 5 3) ๐(๐ฅ ) = √๐ฅ − 1 12 4) ๐(๐ฅ ) = √๐ฅ 2 − 4 13 5) ๐(๐ฅ ) = √๐ฅ 2 − 5๐ฅ + 6 14 Rational Function: General formula: ๐(๐ฅ) ๐(๐ฅ ) = ๐(๐ฅ) , where ๐(๐ฅ )๐๐๐ ๐ (๐ฅ ) are polynomials ๐ (๐ฅ ) ≠ 0. Example: Determine the rational functions from the below functions 1) ๐(๐ฅ ) = 2) ๐(๐ฅ ) = 3) ๐(๐ฅ ) = 4) ๐(๐ฅ ) = 5) ๐(๐ฅ ) = 15 ๐ฅ 2 −7๐ฅ −1 +4 3๐ฅ 2 −4๐ฅ 1 2 ๐ฅ −4๐ฅ 2 2๐ฅ 3 −4๐ฅ 2 +1 √3๐ฅ 3 −4๐ฅ−7 1 2๐ฅ 2 +4๐ฅ 3 5๐ฅ 4 +3๐ฅ 2 −1 1 4 7๐ฅ 3 − ๐ฅ 2√๐ฅ+4๐ฅ 2 3๐ฅ 4 −2๐ฅ โน โน โน โน โน Domain of Rational Functions: ๐ท = ๐ − {๐ง๐๐๐๐ โซ}ู ุงุนุฏุง ุงุตูุงุฑ ุงูู ูุงู โฌ Example: Find the domain of the below functions: 1) ๐(๐ฅ ) = 2) ๐(๐ฅ ) = 16 2๐ฅ 2 +7๐ฅ−4 ๐ฅ−1 7๐ฅ 2 −2๐ฅ 3 +1 ๐ฅ 2 −1 3) ๐(๐ฅ ) = 17 5๐ฅ 2 −4๐ฅ+3 ๐ฅ 2 +๐ฅ−6 Algebraic Functions: All previous functions are algebraic functions. Example: Find the domain of the below functions 1 1) ๐(๐ฅ ) = ๐ฅ−3 + √๐ฅ − 1 18 2) ๐(๐ฅ ) = 19 √๐ฅ−4 √7−๐ฅ Trigonometric Functions: Examples: ๐(๐ฅ ) = sin(2๐ฅ − 1) ๐ (๐ฅ ) = cos(3๐ฅ 2 − 1) ๐(๐ฅ ) = sec(2๐ฅ 3 + 4) ๐(๐ฅ ) = csc(4๐ฅ) ๐(๐ฅ ) = cot(4๐ฅ − 1) ๐(๐ฅ ) = tan(2๐ฅ − 1) Exponential Functions: General formula: ๐ (๐ฅ ) = ๐ ๐ฅ , ๐คโ๐๐๐ ๐ ≠ 1 ๐๐๐ ๐ > 0 Examples: ๐(๐ฅ ) = 4 ๐ฅ ; 1 ๐ฅ ๐(๐ฅ ) = (2) Logarithmic Functions: General formula: ๐ (๐ฅ ) = log ๐ ๐ฅ , ๐คโ๐๐๐ ๐ ≠ 1 ๐๐๐ ๐ > 0 Examples: ๐(๐ฅ ) = log 3 ๐ฅ 20 ; ๐ (๐ฅ ) = log 4 (2๐ฅ 2 − 1) Transformation of functions: Vertical and horizontal shifts: Assume that โ > 0. If ๐ฆ = ๐(๐ฅ) is the graph of a given function, then: 1) The graph of ๐ฆ = ๐ (๐ฅ + โ) is shifted to the left h units. 2) The graph of ๐ฆ = ๐ (๐ฅ − โ) is shifted to the right h units. 3) The graph of ๐ฆ = ๐ (๐ฅ ) + โ is shifted upward h units. 4) The graph of ๐ฆ = ๐ (๐ฅ ) − โ is shifted downward h units. 21 Example: Suppose that ๐ฆ = ๐(๐ฅ) is a given function. Describe the new graph of the below function in words. 1) ๐ฆ = ๐(๐ฅ ) − 3 โบ the graph is shifted downward 3 units. 2) ๐ฆ = ๐(๐ฅ ) + 3 โบ the graph is shifted upward 3 units. 3) ๐ฆ = ๐(๐ฅ − 3) โบ the graph is shifted to the right 3 units 4) ๐ฆ = ๐(๐ฅ + 3) โบ the graph is shifted to the left 3 units 22 5) ๐ฆ = ๐(๐ฅ + 6) + 3 โบ the graph is shifted to the left 6 units and then shifted upward 3 units. 6) ๐ฆ = ๐(๐ฅ − 6) + 3 โบ the graph is shifted to the right 6 units and then shifted upward 3 units. 7) ๐ฆ = ๐(๐ฅ − 6) − 3 โบ the graph is shifted to the right 6 units and then shifted downward 3 units. 8) ๐ฆ = ๐(๐ฅ + 6) − 3 โบ the graph is shifted to the left 6 units and then shifted downward 3 units. 23 Reflecting Transformation: If ๐ฆ = ๐(๐ฅ) is the graph of a given function, then 1) The graph of ๐ฆ = −๐(๐ฅ) is reflected the graph about x-axis 2) The graph of ๐ฆ = ๐(−๐ฅ) is reflected the graph about y-axis 24 Example: Suppose that ๐ฆ = ๐(๐ฅ) is a given graph of the function. Describe in words the following new functions 1) ๐ฆ = ๐(−๐ฅ ) โบ the graph is reflected about y-axis 2) ๐ฆ = −๐ (๐ฅ ) โบ the graph is reflected about x-axis 3) ๐ฆ = −๐ (๐ฅ ) + 6 โบ the graph is reflected about x-axis and then shifted upward 6 units. 4) ๐ฆ = ๐(−๐ฅ ) − 6 โบ the graph is reflected about y-axis and then shifted downward 6 units. 25 Arithmetic Combination of functions: Suppose that ๐(๐ฅ ) ๐๐๐ ๐(๐ฅ) are functions, then 1) (๐ + ๐)(๐ฅ ) = ๐(๐ฅ ) + ๐(๐ฅ) 2) (๐ − ๐)(๐ฅ ) = ๐(๐ฅ ) − ๐(๐ฅ) 3) (๐. ๐)(๐ฅ ) = ๐(๐ฅ ). ๐(๐ฅ) ๐ ๐(๐ฅ) 4) (๐) (๐ฅ ) = ๐(๐ฅ) , ๐(๐ฅ) ≠ 0 Example: If ๐(๐ฅ ) = ๐ฅ 2 − 1 and ๐(๐ฅ ) = ๐ฅ − 1 , find: 1) (๐ + ๐)(๐ฅ ) = ๐(๐ฅ ) + ๐(๐ฅ) 2) (๐ − ๐)(๐ฅ ) = ๐(๐ฅ ) − ๐(๐ฅ) 26 3) (๐. ๐)(๐ฅ ) = ๐ (๐ฅ ). ๐(๐ฅ) ๐ ๐ (๐ฅ ) ๐ ๐ (๐ฅ ) 4) (๐) (๐ฅ ) = ๐(๐ฅ) 5) (๐ ) (๐ฅ ) = ๐(๐ฅ) 27 Composition of functions: Suppose that ๐(๐ฅ ) ๐๐๐ ๐(๐ฅ ) are given functions. Then, the composite functions is defined as: (๐๐๐)(๐ฅ) = ๐(๐(๐ฅ )) Example: If ๐(๐ฅ ) = ๐ฅ 2 + 1 and ๐(๐ฅ ) = ๐ฅ − 1 , find: 1) (๐๐๐)(๐ฅ) = ๐(๐(๐ฅ )) = ๐(๐ฅ − 1) 2) (๐๐๐)(๐ฅ) = ๐(๐(๐ฅ )) = ๐(๐ฅ 2 + 1) 28 3) (๐๐๐)(๐ฅ) = ๐(๐(๐ฅ )) = ๐(๐ฅ 2 + 1) 4) (๐๐๐)(๐ฅ) = ๐(๐(๐ฅ )) = ๐(๐ฅ − 1) 29 Domain of Composite functions: ๐ท{(๐๐๐)(๐ฅ)} =๐ท(๐๐๐)(๐ฅ) ∩ ๐ท๐(๐ฅ) ๐ท{(๐๐๐)(๐ฅ)} =๐ท(๐๐๐)(๐ฅ) ∩ ๐ท๐(๐ฅ) 30 and Example: If ๐(๐ฅ ) = √๐ฅ − 4 and ๐(๐ฅ ) = ๐ฅ 2 , find domain of (๐๐๐)(๐ฅ) ( ๐ท{(๐๐๐)(๐ฅ)} ) ๐ท{(๐๐๐)(๐ฅ)} =๐ท(๐๐๐)(๐ฅ) ∩ ๐ท๐(๐ฅ) 31 Exponential Function: General formula: ๐ (๐ฅ ) = ๐ ๐ฅ , ๐คโ๐๐๐ ๐ ≠ 1 ๐๐๐ ๐ > 0 Example: Determine the exponential functions of the following functions: a) ๐(๐ฅ ) = 2๐ฅ 2 ๐ฅ b) ๐(๐ฅ ) = (3) c) ๐(๐ฅ ) = (3.2)๐ฅ d) ๐(๐ฅ ) = (√3) ๐ฅ 1 ๐ฅ e) ๐(๐ฅ ) = (− 2) e) ๐(๐ฅ ) = (1)๐ฅ 1 ๐ฅ f) ๐(๐ฅ ) = − (2) 32 Domain of the exponential function: Domain is ๐ = (−∞, ∞) Example: Find the domain of the below functions a) ๐(๐ฅ ) = 3๐ฅ 4 b) ๐(๐ฅ ) = ( )๐ฅ − 5 7 c) ๐(๐ฅ ) = −(√3)๐ฅ + 4 The domain for any exponential function is always R 33 The range of the exponential functions: Let ๐๐๐ and ๐ > 0 and ๐ ≠ 1. Then: If ๐ฆ = ๐ ๐ฅ ± ๐ โน ๐ {๐ฆ} = (±๐, ∞) If ๐ฆ = −๐ ๐ฅ ± ๐ โน ๐ {๐ฆ} = (− ∞, ±๐) Example: Find the range of the below functions: 1) ๐(๐ฅ ) = 2๐ฅ − 3 โน ๐ {๐} = (−3, ∞) 1 ๐ฅ 2) ๐(๐ฅ ) = (2) + 4 โน ๐ {๐} = (+4, ∞) ๐ฅ 3 3 3) ๐(๐ฅ ) = −(√2) + 2 โน ๐ {๐} = (−∞, 2) 3 ๐ฅ 2 5 7 4) ๐(๐ฅ ) = − ( ) − 34 2 โน ๐ {๐} = (−∞, − ) 7 Laws of exponents: let ๐, ๐ ∈ ๐ + , then : ๐ฅ ๐ฆ 1) ๐ ๐ = ๐ ๐ฅ+๐ฆ 3) (๐ ๐ฅ )๐ฆ = ๐ ๐ฅ๐ฆ ๐๐ฅ 2) ๐๐ฆ = ๐ ๐ฅ−๐ฆ , ๐ ≠ 0 4) (๐๐)๐ฅ = ๐ ๐ฅ ๐ ๐ฅ Example: Use the Laws of Exponential to rewrite and simplify the expression. 8−4 a) 2−10 35 b) (๐ฅ 2 )−3 (4๐ฅ −4 )2 Natural Exponential Function: General Formula: ๐ (๐ฅ ) = ๐ ๐ฅ ๐คโ๐๐๐ ๐ ≈ 2.71 Example: Find the domain and range of the below functions. 1) ๐(๐ฅ ) = ๐ ๐ฅ − 3 2) ๐(๐ฅ ) = ๐ ๐ฅ + 3 3) ๐(๐ฅ ) = −๐ ๐ฅ − 3 4) ๐(๐ฅ ) = −๐ ๐ฅ + 3 36 Horizontal line test: We say that the graph of a function is one-to-one if and only if no horizontal line intersects its graph more than once Example: Determine whether the below graphs of the functions are one-to one function or not. 37 1) 2) 3) 4) One-to-One functions: ๐(๐ฅ ) = ๐ฅ ๐ = ๐(๐ฅ ) ๐๐ 1 – 1 ๐๐ข๐๐๐ก๐๐๐ ๐๐ ๐ ๐๐ ๐๐๐ ๐(๐ฅ ) ๐๐ not 1 – 1 ๐๐ข๐๐๐ก๐๐๐ ๐๐ ๐ ๐๐ ๐๐ฃ๐๐ Example: One-to-one functions Not one-to-one functions ๐(๐ฅ ) = ๐ฅ 3 , ๐(๐ฅ ) = ๐ฅ 5 + 3 ๐(๐ฅ ) = ๐ฅ 2 − 5 , ๐ (๐ฅ ) = 5 − ๐ฅ 4 Example: Which one of the following is a one-to-one function. a) 5 – ๐ฅ 4 38 b) ๐ ๐ฅ 6 c) ๐ฅ 3 + 2 d) ln(๐ฅ 2 − 1) Inverse Function: Definition: Example: If ๐(2) = 5 , then ๐ −1 (5) = 39 Cancellation equations: ๐ −1 (๐ (๐ฅ )) = ๐ฅ , ๐๐๐ ๐๐๐ฆ ๐ฅ ∈ ๐ ๐(๐ −1 (๐ฅ )) = ๐ฅ , ๐๐๐ ๐๐๐ฆ ๐ฅ ∈ ๐ Example: tan(tan−1 ๐ฅ) = ๐ฅ tan−1 (tan ๐ฅ) = ๐ฅ log ๐ (๐ ๐ฅ ) = ๐ฅ ๐log๐ ๐ฅ = ๐ฅ ln(๐ ๐ฅ ) = ๐ฅ ๐ ln ๐ฅ = ๐ฅ Finding the inverse of the function: The graph of a function and the graph of its inverse are symmetrical about the line ๐ = ๐ Ex) The graph of ๐ −1 is obtained by reflecting the graph of ๐ about 40 a) origin b) x-axis c) y-axis d) the line ๐ฆ = ๐ฅ Example: Find the inverse function of the functions below: 1) ๐(๐ฅ ) = ๐ฅ 3 + 2 2) ๐(๐ฅ ) = 41 2๐ฅ−4 ๐ฅ Logarithmic Functions: ๐(๐ฅ ) = log ๐ ๐ฅ , ๐คโ๐๐๐ ๐ ≠ 1 ๐๐๐ ๐ > 0 Note that: ๐(๐ฅ ) = ln ๐ฅ is called the Natural Logarithmic Function Laws of Logarithmic: If ๐ฅ, ๐ฆ ∈ ๐ + ๐๐๐ ๐ ∈ ๐ , ๐กโ๐๐ 1) log ๐ (๐ฅ๐ฆ) = log ๐ (๐ฅ) + log ๐ (๐ฆ) ๐ฅ 2) log ๐ (๐ฆ) = log ๐ (๐ฅ) − log ๐ (๐ฆ) 3) log ๐ (๐ฅ ๐ ) = ๐ log ๐ ๐ฅ 4) log ๐ ๐ = 1 , log ๐ 1 = 0 Example: Use the laws of logarithmic to compute log 2 80 − log 2 5 42 Example: Solve the following equations: 1) ๐ 5−3๐ฅ = 10 2) log 3 (2๐ฅ + 3) = 3 43 Inverse Trigonometric Functions: Example: Evaluate 1 1) sin−1 (2) √3 2) cos −1 ( 2 ) √3 3) sin−1 ( 2 ) 44 Example: Find the domain of the function ๐ฆ = sin−1 (3๐ฅ + 1) Solution: 45