Calculus I - Chapter 1: Functions and Models PDF

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An-Najah National University

2021

Dr.Adnan Daraghmeh

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calculus functions mathematics lecture notes

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Calculus I chapter 1 lecture notes cover functions and models. It details sets of numbers, intervals, and includes examples demonstrating inequalities and equations.

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Calculus I Chapter 1: Functions and Models by Dr.Adnan Daraghmeh Department of Mathematics An-Najah National University 2020-2021 1 Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Pa...

Calculus I Chapter 1: Functions and Models by Dr.Adnan Daraghmeh Department of Mathematics An-Najah National University 2020-2021 1 Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 2 of 113 A Preview of Calculus Set of Numbers: a) The natural numbers: N b) The integer numbers: Z c) The rational numbers: Q d) The real numbers: R Types of Intervals : A subset of the real line is called an interval. If a, b ∈ R, a < b, then The finite intervals: a) Closed interval [a, b] = {x : a ≤ x ≤ b} b) Open interval (a, b) = {x : a < x < b} c) Half open/closed interval (a, b] = {x : a < x ≤ b} or [a, b) = {x : a ≤ x < b} The infinite intervals: d) [a, ∞) = {x : x ≥ a} e) (−∞, b) = {x : x < b} f) (−∞, ∞) = R Note: ∞ and −∞ are symbols not a real numbers. The Cartesian Plane : A Cartesian plane is a graph with one x−axis and one y−axis. These two axes are perpendicular to each other. The origin (O) is in the exact center of the graph. Num- bers to the right of the zero on the x−axis are positive; numbers to the left of zero are negative. For the y−axis, numbers below zero are negative and numbers above are positive. Line and slope: The equation of the line passing through the points (x1 , y1 ) and (x2 , y2 ) is: The slope is ∆y y2 − y1 m= = ∆x x2 − x1 The equation is y − y1 = m(x − x1 ) Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 3 of 113 Example 2: Solve the following inequalities and equations a) 3x − 10 > 11. solution: b) 3x2 − 10 = 2. solution: c) 3x2 − 10 ≤ 2. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 4 of 113 Section 1.1: Four Ways to Represent a Function Function: A function f is a rule that assigns to each element x in a set D exactly one element, called f (x), in a set E. The set D is called the domain of the function. The number y = f (x) is the value of f at x and is read f of x. The range of f is the set of all possible values of y = f (x) as x varies throughout the domain. A symbol that represents an arbitrary number in the domain of a function f is called an independent variable. A symbol that represents a number in the range of f is called a dependent variable. Cartesian plane, is divided into four quadrants by two perpendicular lines called the x−axis, a horizontal line, and the y−axis, a vertical line. These axes intersect at a point called the origin. Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 5 of 113 Example 1: State the domain and the range of the following functions Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 6 of 113 The Vertical Line Test: A curve in the xy−plane is the graph of a function of x if and only if no vertical line intersects the curve more than once. The number y = f (x) is the value of f at x and is read f of x. Increasing and Decreasing Functions: A function f is called increasing on an interval I if f (x1 ) < f (x2 ) whenever x1 < x2 It is called decreasing on I if f (x1 ) > f (x2 ) whenever x1 < x2 Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 7 of 113 Example 1: Determine whether the curve is the graph of a function of x. If it is, state the domain and range of the function. Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 8 of 113 Example 2: The graphs of f and g are given. a) Find the domain and the range of f and g. solution: b) State the values of f (−4),f (4), g(3) and g(4). solution: c) On what interval is f decreasing? solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 9 of 113 d) On what interval is g increasing? solution: e) For what value(s) of x is f (x) = g(x)? solution: f ) For what value(s) of x is f (x) ≤ g(x)? solution: g) For what value(s) of x is f (x) ≥ g(x)? solution: h) Solve the equation f (x) = 3, g(x) = 0.5? solution: i) Find the y− and x−intercepts. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 10 of 113 Section 1.2: Mathematical Models: A Catalog of Essential Functions A mathematical model is a mathematical description of a real-world situation. Often the model is a function rule or equation of some type. Polynomials: A function f is called a polynomial if f (x) = an xn + an−1 xn−1 + an−2 xn−2 + · · · + a2 x2 + a1 x + a0 where n is a nonnegative integer and an , an−1 , an−2 , · · · , a0 are constants called the coefficients of the polynomial. If the leading coefficient an 6= 0, then the degree of the polynomial is n. Linear function: A polynomial of degree 1 is of the form f (x) = ax + b, a 6= 0 where a is the slope and (0, b) is the y−intercept (Its graph is always a line). Quadratic function: A polynomial of degree 2 is of the form f (x) = ax2 + bx + c, a 6= 0 Its graph is always a parabola. Cubic function: A polynomial of degree 3 is of the form f (x) = ax3 + bx2 + cx + d, a 6= 0 The domain of any polynomial is R = (−∞, ∞). Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 11 of 113 Example 1: Find a formula of a linear function f such that f passing through the point (2, −3) and f (−5) = 18 solution: Example 2: Sketch the graph of the function f (x) = 0.5x − 3, g(x) = mx + 3 Find the slope, the x−, y−intercepts, the domain and the range. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 12 of 113 Power Functions: A function of the form f (x) = xn where n is a constant, is called a power function. We consider several cases. f (x) = xn , n is a positive integer Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 13 of 113 The general shape of the graph of f (x) = xn is similar to the graph of y = x2 , if n is even. y = x3 , if n is odd. 1 √ f (x) = x n = n x is called a root function, n is a positive integer. Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 14 of 113 √ n The general shape of the graph of f (x) = x is similar to the graph of √ y = x, if n is even. √ y = 3 x, if n is odd. p n The domain of the function f (x) = g(x) If n is odd, then Df = R = (−∞, ∞) If n is even, then Df = {x : g(x) ≥ 0} Example 1: Find the domain of the function. √ a) f (x) = 3 5x − 4 solution: √ b) f (x) = 5x − 4 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 15 of 113 √ c) f (x) = x2 − 5x + 6 solution: √ d) f (x) = 4 x2 − 16 solution: √ e) f (x) = 16 − x2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 16 of 113 The graph of the reciprocal function f (x) = x−1 = 1 x Rational Functions: A rational function f is a ratio of two polynomials: P (x) f (x) = Q(x) where P and Q are polynomials. The domain consists of all values of x such that Q(x) 6= 0. Df = {x ∈ R : Q(x) 6= 0} or Df = R − {x : Q(x) = 0} Example 2: Find the domain of the function. 2x − 3 a) f (x) = x2 − 4x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 17 of 113 x2 − 9 b) f (x) = x2 − 4x + 5 solution: 6x c) f (x) = x2 +4 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 18 of 113 Algebraic Functions: A function f is called an algebraic function if it can be constructed using algebraic operations (such as addition, subtraction, multiplication, division, and taking roots) starting with polynomials. p f (x) = x2 + 1 x+2 g(x) = √ x+ x 4x + 5 √ h(x) = 2 √ + (x − 2) 3 x − 4 x + x Example 3: Find the domain of the function. √ x2 − 4 a) f (x) = 2 x + 6x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 19 of 113 1 b) g(x) = √ 4 x2 − 5x solution: 1 c) h(x) = √ 3 x2− 5x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 20 of 113 √ √ d) f (x) = 8 − 2x + 2x − 6 solution: 3 e) f (p) = p √ 2− p solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 21 of 113 Example 4: Find the domain and the range of the function. √ a) f (x) = 4 − x2 solution: √ b) f (x) = − 4 − x2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 22 of 113 √ c) f (x) = − 4 − x solution: 4 − t2 d) h(x) = 2−t solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 23 of 113 Exponential Functions: The exponential functions are the functions of the form f (x) = ax where the base a is a positive constant. Df = R = (−∞, ∞) Rf = (0, ∞) Laws of Exponents: If a and b are positive numbers and x and y are any real numbers, then a) bx by = bx+y bx b) y = bx−y b c) (bx )y = bxy d) (ab)x = ax bx Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 24 of 113 Logarithmic Functions: The logarithmic functions are the functions of the form f (x) = loga x where the base a is a positive constant. Df = (0, ∞) Rf = R = (−∞, ∞) Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 25 of 113 Example 5: Find the domain of the function. a) f (x) = log2 (x2 − 2x) solution: √ b) f (x) = 9 − x2 + log4 (x − 2) solution: c) h(x) = 5(x−3) + log3 (2 − x) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 26 of 113 Trigonometric Functions: Trigonometric functions can be defined as functions of an angle in terms of the ratio of two of the sides of a right triangle containing the angle. Using ratios of the sides of a right triangle, opposite sine =⇒ sin = hypotenuse adjacent cosine =⇒ cos = hypotenuse opposite sin tangent =⇒ tan = = adjacent cos The sine and cosine functions have domain D = (−∞, ∞) and range R = [−1, 1] and a period of 2π. sin2 x + cos2 x = 1, x ∈ R −1 ≤ sin x ≤ 1, −1 ≤ cos x ≤ 1 sin(x + 2π) = sin x, cos(x + 2π) = cos x π The tangent function has domain D = {x|x ∈ R, x 6= + kπ}, k integer 2 tan(π + x) = tan x, ∀x cos cotangent =⇒ cot = sin 1 secant =⇒ sec = cos 1 csecant =⇒ csc = sin Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 27 of 113 Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 28 of 113 Example 6: Find the domain of the function. cos x a) f (x) = 1 − sin x solution: 1 b) f (x) = 1 − 2 cos x solution: 1 c) h(x) = 1 − tan x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 29 of 113 Example 7: Classify each function as a power function, root function, polynomial (state its degree), rational function, algebraic function, trigonometric function, exponential function, or logarithmic function. √ a) g(x) = 4 x solution: 2x3 b) f (x) = 1 − x2 solution: c) w(θ) = sin θ cos2 θ solution: d) y = tan t − cos t solution: e) y = x2 (2 − x3 ) solution: √ x3 − 1 f) y = √ 1− 3x solution: g) y = π x solution: h) y = xπ solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 30 of 113 Example 8: a) Find an equation for the family of linear functions with slope 2 and sketch several members of the family. solution: b) Find an equation for the family of linear functions such that f (2) = 1 and sketch several members of the family. solution: c) Which function belongs to both families? solution: d) What do all members of the family of linear functions f (x) = 1 + m(x + 3) have in common? Sketch several members of the family. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 31 of 113 Piecewise Defined Functions: The functions are defined by different formulas in different parts of their domains. Example 9: Sketch the graph of the function f and find its domain and range ( −3x − 8, −4 ≤ x ≤ 0 f (x) = −3x + 9, 0 < x ≤ 3 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 32 of 113 Example 10: Sketch the graph of the function f and find its domain and range ( x − 1, x ≥ 2 f (x) = x2 , 2 0. To obtain the graph of a) y = f (x) + c, shift the graph of y = f (x) a distance c units upward b) y = f (x) − c, shift the graph of y = f (x) a distance c units downward c) y = f (x − c), shift the graph of y = f (x) a distance c units to the right d) y = f (x + c), shift the graph of y = f (x) a distance c units to the left Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 47 of 113 Transformations of Functions: Vertical and Horizontal Stretching and Reflecting : Suppose c > 1. To obtain the graph of a) y = cf (x), stretch the graph of y = f (x) vertically by a factor of c 1 b) y = f (x), shrink the graph of y = f (x) vertically by a factor of c c c) y = f (cx), shrink the graph of y = f (x) vertically by a factor of c x d) y = f ( ), stretch the graph of y = f (x) horizontally by a factor of c c e) y = −f (x), reflect the graph of y = f (x) about the x−axis f ) y = f (−x), reflect the graph of y = f (x) about the y−axis Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 48 of 113 √ Example 1: Given the graph of y = x, use transformations to graph √ a) f (x) = x − 2 solution: √ b) f (x) = x + 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 49 of 113 √ c) f (x) = x − 2 solution: √ d) f (x) = x + 2 solution: √ e) f (x) = 2 − x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 50 of 113 √ f ) f (x) = 3 x solution: √ g) f (x) = 21 x solution: √ h) f (x) = 3x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 51 of 113 q i) f (x) = 12 x solution: √ j) f (x) = − −x solution: √ k) f (x) = x − 2 − 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 52 of 113 Example 2: Given the graph of y = x2 , use transformations to graph a) f (x) = −x2 + 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 53 of 113 b) f (x) = 4x2 solution: c) f (x) = (2x − 1)2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 54 of 113 d) f (x) = 2x2 + 12x + 20 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 55 of 113 Example 3: Given the graph of y = |x|, use transformations to graph a) f (x) = −2|x| + 3 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 56 of 113 b) f (x) = 3|x − 2| + 1 solution: c) f (x) = |x2 − 1| solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 57 of 113 Example 4: Find expressions for the quadratic functions whose graph are shown. Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 58 of 113 Example 5: Sketch the graphs of the following functions. a) f (x) = sin 2x solution: b) f (x) = 1 − sin 2x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 59 of 113 c) f (x) = cos 21 x solution: d) f (x) = cos 2x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 60 of 113 Example 6: a) How is the graph of y = f (|x|) related to the graph of f ? solution: b) Sketch the graph of y = sin|x| solution: p c) Sketch the graph of y = |x| solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 61 of 113 Example 7: The graph of f is given. Draw the graphs of the following functions. a) y = f (x) + 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 62 of 113 b) y = 2f (x) − 1 solution: c) y = 12 f (x) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 63 of 113 d) y = f ( 12 x) solution: e) y = f (2x) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 64 of 113 f ) y = f (x − 1) solution: g) y = −f (−x) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 65 of 113 √ Example 8: The graph of y = 3x − x2 is given. Use transformations to create a function whose graph is as shown. a) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 66 of 113 b) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 67 of 113 Example 9: Suppose f is a function with domain [−3, 7] and range [1, 8]. Find the domain and the range of the following functions a) g(x) = 6f (2x − 1) + 5 solution: b) h(x) = 3f (2x + 1) − 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 68 of 113 Combinations of Functions: Let f and g be two functions, then:   a) f ∓ g (x) = f (x) ∓ g(x) and domain f ∓ g is Df ∓g = Df ∩ Dg   b) f × g (x) = f (x) × g(x) and domain f × g is Df ×g = Df ∩ Dg f  f (x) f c) (x) = and domain is g g(x) g D f = {Df ∩ Dg | g(x) 6= 0} g Definition: Given two functions f and g, the composite function f ◦ g (also called the composition of f and g) is defined by     f ◦ g (x) = f g(x)   The domain of f ◦ g is the set of all x in the domain of g such that f g(x) is in the domain of f. Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 69 of 113 √ √ Example 1: Given f (x) = x and g(x) = 3 − x. Find a formula for each of the following functions and state their domains.   a) f + g (x) = solution:   b) f × g (x) = solution: f  c) (x) = g solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 70 of 113   d) f ◦ g (x) = solution:   e) g ◦ f (x) = solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 71 of 113 √ 3   Example 2: Given k(x) = x − 2 and h(x) =. Find h◦k (x) and state its domain. 4 − x2 Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 72 of 113 Example 3: Use the given graphs of f and g to evaluate each expression, or explain why it is undefined.   a) f ◦ g (0) solution:   b) f ◦ f (4) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 73 of 113   c) g ◦ f (6) solution:   d) g ◦ g (−2) solution:   e) f g(2) solution:   f ) g f (0) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 74 of 113 Example 4: If g(x) = 2x + 1 and h(x) = 4x2 + 4x + 7, find a function f such that f ◦ g = h. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 75 of 113 Example 5: If f (x) = 3x + 5 and h(x) = 3x2 + 3x + 2, find a function g such that f ◦ g = h. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 76 of 113 Example 6: Find functions f , g, and h such that F = f ◦ g ◦ h (or Express the function in the form f ◦ g ◦ h) a) F (x) = cos2 (x + 9) solution: p b) F (x) = 8 2 + |x| solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 77 of 113 Example 7: Suppose g is an odd function and let h = f ◦ g. Is h always an odd function? What if f is odd? What if f is even? solution: Example 8: Suppose g is an odd function and f is an even function such that f (3) = g(3) = 7, f (7) = −4. Find the value of (3f − 2g)(−3) and (f ◦ g)(−3) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 78 of 113   x Example 9: If f ◦ g (x) = x and f (x) =. Find g(x) x−1 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 79 of 113 Section 1.4: Exponential Functions The Number e: e is the number such that e ≈ 2.71828 The Natural Exponential Function: The natural exponential function is defined as f (x) = ex Example 1: Use graph of y = 2x to sketch the graph of the following functions: a) y = 3 − 2x solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 80 of 113 b) y = 2|x| solution: 1 c) y = ( )|x| 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 81 of 113 d) y = |1 − 2x−1 | solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 82 of 113 1 Example 2: Graph the function y = e−x − 1 and state the domain and range. 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 83 of 113 Example 3: Find the exponential function f (x) = c ax whose graph is given. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 84 of 113 Section 1.5: Inverse Functions and Logarithms Definition: A function f is called a one-to-one function (1 − 1) if it never takes on the same value twice; that is, f (x1 ) 6= f (x2 ) whenever x1 6= x2 Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more than once. Example 1: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 85 of 113 The Inverse function: Definition: Let f be a one-to-one function with domain A and range B. Then its inverse function f −1 has domain B and range A and is defined by f −1 (y) = x ⇔ y = f (x) for any y in B. domain of f −1 = range of f range of f −1 = domain of f Note that: 1 1 f −1 (x) 6= and [f (x)]−1 = f (x) f (x) Example 2: Given that f (2) = 5, f (−3) = 4, find the value of 6f −1 (4) − 7f −1 (5) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 86 of 113  −1 Example 3: Use the following table to find f ◦ g (1) x 0 1 2 3 f (x) 1 3 2 0 g(x) 2 4 1 0 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 87 of 113 How to Find the Inverse Function of a One-to-One Function f : 1. Write y = f (x) 2. Solve this equation for x in terms of y (if possible). 3. To express f −1 as a function of x, interchange x and y. The resulting equation is y = f −1 (x). The graph of f −1 is obtained by reflecting the graph of f about the line y = x. Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 88 of 113 Example 1: Use the graph of the function y = f (x) to sketch the graph of a) y = f −1 (x) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 89 of 113 b) y = −f −1 (x + 1) solution: c) y = f −1 (x − 1) + 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 90 of 113 Example 2: Find the inverse function of a) f (x) = 5x3 + 1 solution: √ b) g(x) = 1 − x2 , 0≤x≤1 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 91 of 113 3x − 1 c) f (x) = 5 − 2x solution: d) f (x) = x2 − 2x + 2, x≤1 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 92 of 113 √ Example 3: Sketch the graphs of f (x) = −1 − x and its inverse function using the same coordinate axes. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 93 of 113 Logarithmic Functions: The logarithmic functions defined as y = loga x, a > 0, x > 0 The natural logarithmic functions defined as y = loge x = ln x, x>0 y = loga x ⇐⇒ x = ay y = ln x ⇐⇒ x = ey loga ax = x, x∈R and aloga x = x, x>0 ln ex = x, x∈R and eln x = x, x>0 Laws of Logarithms: If x and y are positive numbers, then 1. loga (xy) = loga x + loga y and ln(xy) = ln x + ln y x x 2. loga = loga x − loga y and ln = ln x − ln y y y 3. loga (xr ) = r loga x and ln(xr ) = r ln x Change of Base Formula: For any positive number a(a 6= 1), we have ln x loga x = ln a Example 1: Express the given quantity as a single logarithm. a) ln b + 2 ln c − 3 ln d solution: 1 1h 2 i b) ln(x + 2)3 + ln x − ln x2 + 3x + 2 3 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 94 of 113 Example 2: Find the exact value of each expression. a) log 40 + log 2.5 = solution: b) log8 60 − log8 3 − log8 5 = solution: 1 c) ln 3 = e solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 95 of 113 Example 3: Solve each equation for x. a) e7−4x = 6 solution: b) 25−x = 3 solution: c) ln x2 − 1 = 3  solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 96 of 113 d) ln x + ln(x − 1) = 1 solution: e) ln(ln x) = 1 solution: f ) log(x−2) 4 = 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 97 of 113 g) e2x − 3ex + 2 = 0 solution: √ h) e 2+x = ex solution: 2 i) 2(x +1) − 4(3x) = 8(2x) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 98 of 113 Example 4: Solve each inequality for x. a) 1 < e3x−1 < 2 solution: b) 1 − 2 ln x < 3 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 99 of 113 Example 5: Find the domain of the function and its inverse a) f (x) = ln(ex − 3) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 100 of 113 ex b) f (x) = ex + 2 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 101 of 113 Example 6: a) If f (x) = 2x + ln x, find f −1 (2). solution: b) If f (x) = x5 + x3 + x, find f −1 (3) and f (f −1 (2)). solution: c) If f (x) = 3 + x + ex , find f −1 (4). solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 102 of 113 Example 7: Sketch the graph of the function f (x) = ln(x − 2) − 1. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 103 of 113 Inverse Trigonometric Functions The inverse sine function or the arcsin function. π π y = arcsin x = sin−1 x ⇐⇒ sin y = x, −1 ≤ x ≤ 1, − ≤y≤ 2 2   π π sin−1 sin x = x, − ≤x≤   2 2 sin sin−1 x = x, −1 ≤ x ≤ 1 Example 1: Evaluate a) sin−1 ( 12 ). solution:  √  b) arcsin −2 2. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 104 of 113   c) sin sin−1 −0.5. solution:   d) sin sin−1 −1. solution:   e) sin sin−1 −1.1. solution:  2π  f ) sin−1 sin. 3 solution:  7π  g) sin−1 sin. 4 solution:  −5π  h) sin−1 sin. 3 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 105 of 113 Example 2: Find the domain and range of the function g(x) = sin−1 (3x + 1) solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 106 of 113 The inverse cosine function or the arccos function. y = arccos x = cos−1 x ⇐⇒ cos y = x, −1 ≤ x ≤ 1, 0≤y≤π   cos−1 cos x = x, 0≤x≤π   cos cos−1 x = x, −1 ≤ x ≤ 1 Example 1: Evaluate −1 a) cos−1 ( √ 2 ). solution:  √  b) arccos −2 2. solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 107 of 113   c) cos cos−1 −0.5. solution:   d) cos cos−1 −1. solution:   e) cos cos−1 −1.1. solution:  2π  f ) cos−1 cos. 3 solution:  7π  g) cos−1 cos. 4 solution:  −5π  h) cos−1 cos. 3 solution: Clculus I, 2021 Dr.Adnan Daraghmeh Lecture Notes - Page 108 of 113 Example 2: Find the domain and range of the function g(x) = 5 + 2 cos−1 (3x + 1) solution: The inverse tangent function or the arctan function. −π π y = arctan x = tan−1 x ⇐⇒ tan y = x, x ∈ R,

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