1.5 Inverse Trigonometric Functions

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1 1.5 Inverse Trigonometric Functions Remember that only one-to-one functions have inverses. So, in order to find the inverse functions for sine, cosine, and tangent, we must restrict their domains to intervals where they are one-to-one. To find the inverse sine function, we restrict the domain of sine to [ π/2, π/2]. sin 1 x = y sin y = x In other words, sin 1 x or arcsin x is the ANGLE in the interval [ π/2, π/2] whose sine is x. sin arcsin( 1 2 ) sin 1 2 In order to have an inverse for cosine, we restrict the domain of cosine to the interval [0, π]. cos 1 x = y cos y = x In other words, cos 1 x or arccos x is the ANGLE in the interval [0, π] whose cosine is x. ( arccos(0) cos 1 ) 2 2 arccos 1 2 ( ) arccos 5 4 1

2 In order to have an inverse for tangent, we restrict the domain of tangent to the interval ( π/2, π/2). tan 1 x = y tan y = x y = tan x y = arctan x In other words, tan 1 x or arctan x is the ANGLE in the interval ( π/2, π/2) whose tangent is x. arctan(1) tan 1 ( 3) When combining trig and inverse trig, remember that an inverse trig expression is an ANGLE!! tan(sin (cos ( )) ) sin sin(tan 1 x) cot(cos 1 x) 2

3 2.2 The Limit of a Function Introductory Example: Consider the function f(x) = function when x is near 0. x f(x) x f(x) x x The table below gives values of the As the values of x get closer to 0 from both sides, the values of f(x) are getting closer to 4. This is written x mathematically as: = 4 x 0 x f(x) = L x a is read as the it as x approaches a of f(x) is L. This means the values of f(x) get closer and closer to L (or possibly equal to L) by letting x get closer and closer to a, but not equal to a. 0 f(0) is undefined in the above example since f(0) = = 0 0. When dealing with its, we are examining values as x approaches a, but not equal to a. [Note: We ll see later that the value of the function at a may or may not equal the it.] In all three of these pictures, x 4 f(x) = 2. It doesn t matter what f(4) actually is or even if it exists. Left-Handed Limit: f(x) x a i.e. values of x less than a. is the it when ONLY looking at values of x approaching from the left, Right-Handed Limit: f(x) is the it ONLY looking at values of x approaching from the right, i.e. x a + values of x greather than a. The it exists if and only if the left-handed and right-handed its both exist and are equal. f(x) = L if and only if f(x) = L and f(x) = L x a x a x a + 3

4 1 Infinite Limits: Calculate x 0 x 2 x f(x) x f(x) , , , 000, , 000, , 000, , 000, 000 As the values of x approach 0 from both sides, f(x) gets larger and larger without bound. In this case, we 1 say that x 0 x 2 =. 1 Example: Calculate x 0 + x, x x 0 1 1, and x 0 x Definition: The line x = a is a vertical asymptote if the it from the left, right, or both is or. Example: Consider the graph of f below. Find the indicated its. f(x) x 4 f(x) x 4 + f(x) x 4 f(x) x 2 f(x) x 2 + f(x) x f(x) x 1 4 f(x) x 6 2 f(x) x 3 f(x) x 3 + f(x) x f(x) x What are the vertical asymptotes of f(x)? 8 4

5 When the it of a function at x = a is of the form nonzero 0, then there is a vertical asymptote at x = a and the it will be either or. Examples: x 1 x 5 x + 5 x 1 x 5 + x + 5 x 1 x 5 x + 5 x + 3 x 2 2 x x 3 x 3 (x 6)(x 3) 2 ln x x 0 + x 4 ln(x3 64) + Vertical Asymptotes vs Holes Be careful with rational functions. Just because there is division by zero does NOT mean there is a vertical asymptote. If a factor cancels completely from the denominator, then there is a hole there, not a vertical asymptote, because the division by 0 is fixed. Find all vertical asymptotes of the function f(x) = x 2 x 2 6x

6 2.3 Calculating Limits Using the Limit Laws Limit Laws: Suppose x a f(x) and x a g(x) exist and that c is any constant. 1. x a (f(x) ± g(x)) = x a f(x) ± x a g(x) 2. x a cf(x) = c x a f(x) 3. x a f(x)g(x) = x a f(x) x a g(x) f(x) 4. x a g(x) = f(x) x a x a 5. x a (f(x)) n = g(x) provided that g(x) 0 x a ( ) n f(x) x a 6. x a n f(x) = n x a f(x). If n is even, then we must have that x a f(x) > 0 If f is not a piecewise function and a is in the domain of f, then x a f(x) = f(a). In other words, if you can evaluate the function at a and you don t get division by 0 or something undefined, then that value IS the it! x 2 (3x2 + 5x + 1) 4 Given that x 3 f(x) = 16, calculate x 3 (x 2 4)f(x) f(x) + x + 2 6

7 When the it is of the form 0 0 we say the it is indeterminate. Two things could be happening. There is either a vertical asymptote or a hole in the graph at x = a. If after using algebra, the it simplifies to nonzero 0, then there is a vertical asymptote. If the it simplifies to an actual number, then there is a hole. So, if you get a it of the form 0 0, you must USE ALGEBRA to determine the it. Some methods used are expanding, factoring, or multiplying by the conjugate of a radical. x 2 x 12 x 4 x 2 16 (h 4) 2 16 h 0 h t 2 t 2 (t 2) 3 x 1 x x 1 7

8 For vector functions, if r(t) = f(t), g(t), then r(t) = f(t), t a t a g(t) t a provided the its of the component functions exist. 9t 1 3(t 2) 1 Calculate r(t), where r(t) =, t 3 t 3 t 3 t

9 Let f(x) = x if x < 0 x 2 if 0 x < 2 8 x if 2 x < 5 2 if x = 5 x 2 if x > 5 Calculate x 0 f(x), x 2 f(x), and x 5 f(x) or explain why the it does not exist. Recall the definition of the absolute value function: { { x = x + 4 = Calculate x 0 x 2 + x x or explain why the it does not exist. 9

10 Calculate x 3 x 3 6 2x or explain why the it does not exist. Squeeze Theorem: If g(x) h(x) f(x) for all x in an interval that contains a (except possibly at a) and g(x) = f(x) = L x a x a then x a h(x) = L Example: If 4x 2 f(x) x for 0 x 3, find x 2 f(x). ( ) Example: Find x 2 sin 1 x 0 x. 10

11 2.5 Continuity In Section 2.3 we saw that the it as x approaches a can sometimes be found by evaluating the function at a. If this is the case, then the function is continuous. Definition: A function is continuous at a number a if f(x) = f(a) x a Otherwise, we say the function is discontinuous at a, or that there is a discontinuity at a. In order for a function to be continuous at a number a: (1) f(a) must be defined. So a function will NOT be continuous anywhere it is undefined. (2) x a f(x) must exist. (The left-handed and right-handed its must both equal the same value.) (3) x a f(x) = f(a) All polynomials are continuous everywhere! Rational functions are continuous wherever they are defined, i.e. where the denominator is not 0. Examples of discontinuities: Holes, vertical asymptotes, and jumps. A hole in a graph is also referred to as a removable discontinuity because if we wanted to, we could just redefine the function at that point to make it continuous. Removable discontinuities occur where the it exists at a (left and right its are equal), but is not equal to f(a). A vertical asymptote is referred to as an infinite discontinuity. A jump in the graph is referred to as a jump discontinuity. Jumps occur where the its from the left and right exist, but are not equal. 11

12 A function is continuous from the left at a number a if f(x) = f(a) and continuous from the x a right if f(x) = f(a). A function is continuous if and only if it is continuous from both the right and x a + the left. Examples: Determine where the functions below are discontinuous. Explain why mathematically. Is the function continuous from the left or right at any discontinuity? f(x) = x 2 25 (x 5)(x + 3) { x f(x) = 2 4 if x 1 x + 1 if x > 1 2x 1 if x < 4 f(x) = 6 if x = 4 x 2 9 if x > 4 12

13 f(x) = 3x + 1 x 2 5 x 1 x 3 25 x 2 if x < 2 if 2 x 3 if x > 3 What values of a and b would make the following function continuous everywhere? ax 2 + bx + 1 if x 3 f(x) = x 2 b if 3 < x < 1 ax + 5b if x 1 13

14 The Intermediate Value Theorem: Suppose f is continuous on the closed interval [a, b] and let N be any number strictly between f(a) and f(b). Then there exists a number c in (a, b) such that f(c) = N. Example: Show that the equation x 3 + 2x + 2 = 0 has a root (solution) on the interval (1, 2). Example: If f(x) = x 4 x 3 + 3x 2 + 2, show that there is a number c so that f(c) = 3. 14