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. √ −1 3 1 −1 sin 2 arcsin(− 2 ) sin 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. √ −1 2 1 5 arccos(0) cos − 2 arccos 2 arccos − 4
1 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!!
−1 4 −1 2 tan(sin 5 ) sin cos − 3
sin(tan−1 x) cot(cos−1 x)
2 2.2 The Limit of a Function x Introductory Example: Consider the function f(x) = √ . The table below gives values of the x + 4 − 2 function when x is near 0. x f(x) x f(x) −0.5 3.8708287 0.5 4.1213203 −0.1 3.9748418 0.1 4.0248457 −0.05 3.9874607 0.05 4.0124612 −0.01 3.9974984 0.01 4.0024984 −0.001 3.999750 0.001 4.000250 −0.0001 3.999975 0.0001 4.000025 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: lim √ = 4 x→0 x + 4 − 2 lim f(x) = L x→a is read as “the limit 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. f(0) is undefined in the above example since f(0) = √ 0 = 0 . When dealing with limits, we are 0+4−2 0 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 limit.] In all three of these pictures, lim f(x) = 2. It doesn’t matter what f(4) actually is or even if it exists. x→4
Left-Handed Limit: lim f(x) is the limit when ONLY looking at values of x approaching from the left, x→a− i.e. values of x less than a. Right-Handed Limit: lim f(x) is the limit ONLY looking at values of x approaching from the right, i.e. x→a+ values of x greather than a.
The limit exists if and only if the left-handed and right-handed limits both exist and are equal. lim f(x) = L if and only if lim f(x) = L and lim f(x) = L x→a x→a− x→a+
3 1 Infinite Limits: Calculate lim x→0 x2 x f(x) x f(x) −0.5 4 0.5 4 −0.1 100 0.1 100 −0.05 400 0.05 400 −0.01 10, 000 0.01 10, 000 −0.001 1, 000, 000 0.001 1, 000, 000 −0.0001 100, 000, 000 0.0001 100, 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 lim = ∞. x→0 x2 1 1 1 Example: Calculate lim , lim , and lim x→0+ x x→0− x x→0 x
Definition: The line x = a is a vertical asymptote if the limit from the left, right, or both is ∞ or −∞.
Example: Consider the graph of f below. Find the indicated limits. lim f(x) lim f(x) lim f(x) x→4− x→4+ x→4 8 lim f(x) lim f(x) lim f(x) − + x→2 x→2 x→2 6
lim f(x) 4 x→−1
2 lim f(x) x→−6
−8 −6 −4 −2 2 4 6 8 lim f(x) lim f(x) lim f(x) x→−3− x→−3+ x→−3 −2
lim f(x) −4 x→6
−6
What are the vertical asymptotes of f(x)? −8
4 nonzero When the limit of a function at x = a is of the form 0 , then there is a vertical asymptote at x = a and the limit will be either ∞ or −∞. Examples: x − 1 x − 1 x − 1 lim lim lim x→−5− x + 5 x→−5+ x + 5 x→−5 x + 5
x + 3 x3 lim lim x→2− 2 − x x→3 (x − 6)(x − 3)2
lim ln x lim ln(x3 − 64) x→0+ x→4+
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.” x − 2 Find all vertical asymptotes of the function f(x) = . x2 − 6x + 8
5 2.3 Calculating Limits Using the Limit Laws Limit Laws: Suppose lim f(x) and lim g(x) exist and that c is any constant. x→a x→a
1. lim(f(x) ± g(x)) = lim f(x) ± lim g(x) x→a x→a x→a 2. lim cf(x) = c lim f(x) x→a x→a 3. lim f(x)g(x) = lim f(x) lim g(x) x→a x→a x→a
lim f(x) 4. lim f(x) = x→a provided that lim g(x) 6= 0 x→a g(x) lim g(x) x→a x→a n 5. lim(f(x))n = lim f(x) x→a x→a
pn q 6. lim f(x) = n lim f(x). If n is even, then we must have that lim f(x) > 0 x→a x→a x→a
If f is not a piecewise function and a is in the domain of f, then lim f(x) = f(a). In other words, if you x→a can evaluate the function at a and you don’t get division by 0 or something undefined, then that value IS the limit! lim (3x2 + 5x + 1)4 x→−2
p(x2 − 4)f(x) Given that lim f(x) = 16, calculate lim x→3 x→3 f(x) + x + 2
6 0 When the limit is of the form 0 we say the limit 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 limit simplifies to nonzero 0 , then there is a vertical asymptote. If the limit simplifies to an actual number, then there is a hole. 0 So, if you get a limit of the form 0 , you must USE ALGEBRA to determine the limit. Some methods used are expanding, factoring, or multiplying by the conjugate of a radical. x2 − x − 12 lim x→4 x2 − 16
(h − 4)2 − 16 lim h→0 h
t − 2 lim t→2 (t − 2)3
√ x + 3 − 2 lim x→1 x − 1
7 For vector functions, if r(t) = hf(t), g(t)i, then lim r(t) = lim f(t), lim g(t) t→a t→a t→a provided the limits of the component functions exist. * + 9t−1 − 3(t − 2)−1 t − 3 Calculate lim r(t), where r(t) = , √ t→3 t − 3 t2 + 7 − 4
8 x if x < 0 x2 if 0 ≤ x < 2 Let f(x) = 8 − x if 2 ≤ x < 5 −2 if x = 5 x − 2 if x > 5 Calculate lim f(x), lim f(x), and lim f(x) or explain why the limit does not exist. x→0 x→2 x→5
Recall the definition of the absolute value function: ( ( |x| = |x + 4| =
x2 + x Calculate lim or explain why the limit does not exist. x→0 |x|
9 |x − 3| Calculate lim or explain why the limit does not exist. x→3 6 − 2x
Squeeze Theorem: If g(x) ≤ h(x) ≤ f(x) for all x in an interval that contains a (except possibly at a) and
lim g(x) = lim f(x) = L x→a x→a then lim h(x) = L x→a
Example: If 4x − 2 ≤ f(x) ≤ x2 + 2 for 0 ≤ x ≤ 3, find lim f(x). x→2
Example: Find lim x2 sin 1 . x→0 x
10 2.5 Continuity In Section 2.3 we saw that the limit 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
lim 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) lim f(x) must exist. (The left-handed and right-handed limits must both equal the same value.) x→a (3) lim f(x) = f(a) x→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 limit exists at a (left and right limits 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 limits from the left and right exist, but are not equal.
11 A function is continuous from the left at a number a if lim f(x) = f(a) and continuous from the x→a− right if lim 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? x2 − 25 f(x) = (x − 5)(x + 3)
( x2 − 4 if x ≤ −1 f(x) = x + 1 if x > −1
2x − 1 if x < 4 f(x) = 6 if x = 4 x2 − 9 if x > 4
12 3x + 1 if x < −2 x2 − 5 if − 2 ≤ x ≤ 3 f(x) = x − 1 3 x − 25 if x > 3 x − 2
What values of a and b would make the following function continuous everywhere? ax2 + bx + 1 if x ≤ −3 f(x) = x2 − b if − 3 < x < 1 ax + 5b if x ≥ 1
13 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 −x3 + 2x + 2 = 0 has a root (solution) on the interval (1, 2).
Example: If f(x) = x4 − x3 + 3x2 + 2, show that there is a number c so that f(c) = 3.
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