UNIT CIRCLE, Differential Calculus

Domain and range of f(x) = 1/2+e^4x

Domain: (-∞,∞) Range: (0, 1/2)

Domain and Range y = sin⁻¹x

Domain: -1 ≤ x≤1 Range: -π/2 ≤y ≤π/2 Alternate notation: sin⁻¹x = arcsin x

Domain and Range y = cos⁻¹x

Domain: -1≤x≤1 Range: 0≤y≤π Alternate Notation: cos⁻¹x = arccos x

Domain and Range y = tan⁻¹x

Domain: -∞<x<∞ Range: -π/2<y<π/2 Alternate Notation: tan⁻¹x = arctan x

Domain of Secx: Range of secx:

Domain: x ≠ π/2 +nπ Range: (-∞,-1)∪[1,∞)

Range: The range is all possible values to get out of the function. sec θ =

sec θ ≥ 1 and sec θ ≤ -1

Pythagorean Identities tan²θ + 1 =

sec²θ

Even/Odd Formulas sec(-θ)

secθ

To find the vertical asymptote(s) of a rational function,

simply set the denominator equal to 0 and solve for x.

Periodic Formulas sin ( θ + 2πn) =

sin θ

Tangent and Cotangent Identities tan θ =

sin θ/cos θ

Periodic Formulas tan ( θ + πn) =

tan θ

ln of e

1

lim |x − 2|/x − 2 . x→2

Left limit does not equal to the right limit. The limit does not exist, and also not ±∞.

Two ways you can find range

Graph function Find domain of inverse

Sketch the function f(x) = 1 + |x − 2|. What is the range of f?

Horizontal shift right by 2 • Vertical shift up by 1 Range: [1,∞)

Unit Circle

The circle of radius one centered at the origin in the xy-plane.

1. f(x)=|x| range

The range of f(x) is [0,∞), which is all non-negative real numbers.

e raised to the ln power

x

ln of e raised to the x power

x

Calculating the solutions for cosx

x = α + 2nπ or x = (2π-a) + 2nπ

Calculating the solutions for Sinx

x = α + 2nπ or x = (π-a) + 2nπ

If f = 1/(x − 2) and g =√x, construct expressions for f ◦ g and for g ◦ f and find their domains.

√ x − 2 ≠ 0 √ x ≠2 x ≠4 so the domain is all real numbers that are greater than or equal to 0, excluding 4. That is [0,4)∪(4,∞).

30º

π/6 (√3/2, 1/2)

cos(180°)

−1

cos(π)

−1

sin(270°)

−1

sin(3π/2)

−1

cos(120°)

−1/2

cos(240°)

−1/2

cos(2π/3)

−1/2

cos(4π/3)

−1/2

sin(11π/6)

−1/2

sin(210°)

−1/2

sin(330°)

−1/2

sin(7π/6)

−1/2

cos(135°)

−√2/2

cos(225°)

−√2/2

cos(3π/4)

−√2/2

cos(5π/4)

−√2/2

sin(225°)

−√2/2

sin(315°)

−√2/2

sin(5π/4)

−√2/2

sin(7π/4)

−√2/2

cos(150°)

−√3/2

cos(210°)

−√3/2

cos(5π/6)

−√3/2

cos(7π/6)

−√3/2

sin(240°)

−√3/2

sin(300°)

−√3/2

sin(4π/3)

−√3/2

sin(5π/3)

−√3/2

cos(315°)

√2/2

cos(45°)

√2/2

cos(7π/4)

√2/2

cos(π/4)

√2/2

sin(135°)

√2/2

sin(3π/4)

√2/2

sin(45°)

√2/2

sin(π/4)

√2/2

cos(11π/6)

√3/2

cos(30°)

√3/2

cos(330°)

√3/2

cos(π/6)

√3/2

sin(120°)

√3/2

sin(2π/3)

√3/2

sin(60°)

√3/2

sin(π/3)

√3/2

Reciprocal Identities tan θ =

1/cot θ

Reciprocal Identities sin θ

1/csc θ

Reciprocal Identities cos θ =

1/sec θ

Reciprocal Identities csc θ =

1/sin θ

Reciprocal Identities cot θ =

1/tan θ

Unit circle defintion sec θ =

1/x

Unit circle definition csc θ =

1/y

330º

11π/6 (√3/2, -1/2)

The population of a city is 490,000 and is increasing at a rate of 4.5% each year. Approximately when will the population reach 980​,000? ​ (Use a compound growth​ model.)

16 years

360º

2π (1,0)

120º

2π/3 (-1/2, √3/2)

270º

3π/2 (0, -1)

135º

3π/4 (-√2/2, √2/2)

240º

4π/3 (-1/2, -√3/2)

150º

4π/5 (-√3/2, 1/2)

300º

5π/3 (1/2, -√3/2)

225º

5π/4 (-√2/2, -√2/2)

315º

7π/4 (√2/2, -√2/2)

210º

7π/6 (-√3/2, -1/2)

So the natural logarithm of e is equal to one. ln(e) = loge(e) =

= 1

What does it mean for a function f(x) to be continuous at a point x = a in its domain?

A function is continuous at x = a if the output of the function at a is exactly the limit of the function as x → a. That is when lim x→a f(x) = f(a)

How do we compute average speed? How do we compute average velocity?

Average speed = Change in distance/Change in time. Average velocity = Change in position/Change in time.

Range: The range is all possible values to get out of the function. cos θ =

-1≤cosθ ≤1

Range: The range is all possible values to get out of the function. sin θ =

-1≤sinθ≤1

Even/Odd Formulas cot(-θ) =

-cotθ

Even/Odd Formulas csc(-θ) =

-cscθ

Even/Odd Formulas sin(-θ) =

-sinθ

Even/Odd Formulas tan(-θ) =

-tanθ

Range: The range is all possible values to get out of the function. cot θ =

-∞ < cotθ < ∞

Range: The range is all possible values to get out of the function. tan θ =

-∞ < tanθ < ∞

cos(270°)

0

cos(3π/2)

0

cos(90°)

0

cos(π/2)

0

sin θ = lim θ→0

0

sin(0°)

0

sin(180°)

0

sin(2π)

0

sin(360°)

0

sin(π)

0

0º

0 (1,0)

Pythagorean Identities sin²θ + cos²θ =

1

cos θ = lim θ→0

1

cos(0°)

1

cos(2π)

1

cos(360°)

1

lim x→0 sinx/x

1

sin(90°)

1

sin(π/2)

1

The domain of f consists of all numbers that x can have

1) zeros in the denominator 2) negative number under radical 3) physical restrictions to the problem

cos(300°)

1/2

cos(5π/3)

1/2

cos(60°)

1/2

cos(π/3)

1/2

sin(150°)

1/2

sin(30°)

1/2

sin(5π/6)

1/2

sin(π/6)

1/2

Reciprocal Identities sec θ =

1/cos θ

Rules for horizontal asymptotes

Given a rational function (a fraction in which the numerator and denominator are polynomials), we can determine the equation of the horizontal asymptote simply by comparing the numerator to the denominator. We need to start by identifying the degree of the numerator and denominator. The degree is the exponent on the term with the largest exponent. For instance, x3 −2x2 + 6x + 1 is a third-degree polynomial, because x³ is the largest-degree term. 58 The largest-degree term isn't necessarily the first term. If the same polynomial is written as 6x −2x2 + x3 + 1, it's still a third-degree polynomial, because x³ is the largest-degree term. Once we've identified the degree of the numerator and denominator, then we compare them. N < D: If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is given by y = 0. N > D: If the degree of the numerator is greater than the degree of the denominator, then the function doesn't have a horizontal asymptote. N = D: If the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote is given by the ratio of the coefficients on the highest-degree terms.

Squeeze theorem

If f(x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a itself) and if lim x→a f(x) = lim x→a h(x) = L then lim x→a g(x) = L

The following functions are continuous everywhere in their domain:

Polynomials Rational functions Root functions Trigonometric functions Inverse trigonometric functions Exponential functions Logarithmic functions

Point discontinuities

Previously, we learned that a point discontinuity was a single pinpoint of discontinuity in the graph, at which the one-sided limits are equal, and the general limit therefore exists. But, even though the general limit exists, the function is still discontinuous at that point. Now, we're interested in how we can plug that hole in the graph, by redefining the function in such a way that it becomes continuous at that point. To illustrate how easy this is, we'll take the function we used previously in the lesson on point discontinuities

(b) What are the different kinds of discontinuities and how do we describe them using limits?

Removable discontinuity: This occurs when limx→a f(x) 6= f(a). • Jump discontinuity: This occurs when lim x→a+ f(x) and lim x→a− f(x) both exist but are not equal to each other. • Vertical Asymptote: This occurs when at least one of lim x→a+ f(x) or lim x→a− f(x) is either ∞ or −∞. • Wild oscillating: This occurs when lim x→a+ f(x) and lim x→a− f(x) don't even exist. An example of this was f(x) = sin π/x at x = 0.

11. (a) Which of the following maps are maps of functions? Explain your reasoning.

Solution. I and II are functions. III and IV are not functions, because they assign multiple outputs to some inputs. (b) For each function, determine the range and the domain. Solution. For I, the domain is {a,b,c,d,e} and the range is {c}. For II, the domain is {a,b,c,d,e} and the range is {a,b,x,d,z}.

Is the following argument correct or incorrect? Why? Imagine our football team beat Georgia State 41-0. Since we had 0 points at the beginning of the game and 41 at the end, the Intermediate Value Theorem says that we must have had exactly 25 points at some moment in the game.

Solution. The argument is incorrect. The score is not a continuous function of time, so the Intermediate Value Theorem does not apply.

How can you visualize an instantaneous rate of change graphically?

Solution. The instantaneous rate of change of a function f at an input a is the slope of the tangent line at the point (a,f(a)) on the graph of f.

Sketch y(x) = 1 + sin(2x)

Solution. We can get this graph by starting with sin x, doing a horizontal shrink by 2 to get y = sin(2x), and a shift up by 1 to get y = 1 + sin(2x):

Use the definitions of even and odd functions to decide whether each of the following is even, odd, or neither. (a) f(x) = −3x⁴ + 6x² + 1.

Solution. We want to see how f(−x) relates to f(x). Let's simplify f(−x): f(−x) = −3(−x)⁴ + 6(−x)²+ 1 = −3x⁴ + 6x² + 1 This is the same as f(x), so f(x) is even .

g(x) = x³ + x+ 1.

Solution. We want to see how g(−x) relates to g(x). g(−x) = (−x)³ + (−x)+ 1 = −x³ + x²+ 1 This is not the same as g(x), so the function is not even. This is also not the same as −g(x), so the function is not odd. Therefore, g(x) is neither even nor odd. You might be surprised that the degree of g(x) is odd, but it is not an odd function. This is something to keep in mind!

(a) How can you visualize an average rate of change graphically?

The average rate of change of a function f between inputs a and b is the slope of the secant line through the points (a,f(a)) and (b,f(b)).

Which of the following curves represent a function?

The first one y = x² represents a function. The second one y²= x does not because it fails the vertical line test.

What is the range of y(x) = √ x − 1?

The graph of √ x − 1 is obtained by shifting the graph of √ x to the right by 1. This shift does not affect the range of these functions. So the range of √ x − 1 is the same as that of √ x. That is, the range is [0,∞).

ln of infinity

The limit of natural logarithm of infinity, when x approaches infinity is equal to infinity: lim ln(x) = ∞, when x→∞

ln of 1

The natural logarithm of one is zero: ln(1) = 0

ln of 0

The natural logarithm of zero is undefined: ln(0) is undefined

Example 5 Solve cos(3x)=2

This example is designed to remind you of certain properties about sine and cosine. Recall that −1≤cos(θ)≤1−1≤cos⁡(θ)≤1 and −1≤sin(θ)≤1−1≤sin⁡(θ)≤1. Therefore, since cosine will never be greater that 1 it definitely can't be 2. So THERE ARE NO SOLUTIONS to this equation! It is important to remember that not all trig equations will have solutions.

Which of the rates of change (average or instantaneous) are we able to compute at the moment? What is our current way of "computing" the other?

We can so far only compute an average rate of change. We can approximate an instantaneous rate of change by computing the average rate of change over a small interval that includes the point we care about.

What is the domain of y(x) = √ x − 1?

We need to make sure that whatever is inside the square root is greater than or equal to zero. So we need that x − 1 ≥ 0 i.e. x ≥ 1. So the domain is x ≥ 1 or [1,∞].

(a) ln(ln(e^e)) Solution.

We'll use properties of logs: ln(ln(e e )) = ln(e ln e) = ln(e) = 1

Domain of f(x) = √x-1

[1, ∞)

Right triangle definition For this definition we assume that 0 <θ <π/2 or 0° < θ < ° 90 . Cos θ =

adjacent/hypotenuse

Right triangle definition For this definition we assume that 0 <θ <π/2 or 0° < θ < ° 90 . cot θ

adjacent/opposite

60º

π/3 (1/2, √3/2)

Periodic Formulas cos ( θ + 2πn) =

cos θ

Tangent and Cotangent Identities cot θ

cos θ/sin θ

Even/Odd Formulas cos(-θ) =

cosθ

Periodic Formulas: cot (θ + πn) =

cot θ

Range: The range is all possible values to get out of the function. csc θ =

csc θ ≥ 1 and csc θ ≤ -1

Pythagorean Identities 1 + cot² =

csc²θ

Periodic Formulas csc(θ + 2πn) =

cscθ

45º

π/4 (√2/2, √2/2)

e¹ =

e

f is even if f(-x) =

f(x)

f is odd if - -f(x) =

f(x)

Right triangle definition For this definition we assume that 0 <θ <π/2 or 0° < θ < ° 90 . sec θ =

hypotenuse/adjacent

Right triangle definition For this definition we assume that 0 <θ <π/2 or 0° < θ < ° 90 . csc θ =

hypotenuse/opposite

A secant line

is a straight line joining two points on a function. (See below.) It is also equivalent to the average rate of change, or simply the slope between two points. Secant line = Average Rate of Change = Slope

A tangent line

is a straight line that touches a function at only one point. (See above.) The tangent line represents the instantaneous rate of change of the function at that one point. The slope of the tangent line at a point on the function is equal to the derivative of the function at the same point (See below.) Tangent Line = Instantaneous Rate of Change = Derivative

The natural logarithm of a number x is defined as the base e logarithm of x:

ln(x) = loge(x)

So the natural logarithm of e is the base e logarithm of e: ln(e) =

loge(e)

Right triangle definition For this definition we assume that 0 <θ <π/2 or 0° < θ < ° 90 . tan θ =

opposite/adjacent

Right triangle definition For this definition we assume that 0 <θ <π/2 or 0° < θ < ° 90 . Sin θ =

opposite/hypotenuse

Periodic Formulas sec ( θ + 2πn) =

sec θ

Intermediate value theorem

without an interval When we're asked to use the intermediate value theorem to prove that the function has a root, but we're not given the interval in which to find the root, then we're forced to come up with our own interval. Because the intermediate value theorem can only be used on closed intervals [a, b], we'll need to find a closed interval to investigate. There are different ways we can go about finding the interval, the first of which is to blindly guess. We can try picking two random values to be the ends of the interval, and we might get lucky and find that a negative value at one end and a positive value at the other. If that doesn't work, we might be able to consider some other aspect of the function. For instance, if we're dealing with the trigonometric function sin x or cos x, we know that those functions oscillate back and forth between −1 and 1. So we could choose a value where the trig function is −1, and another value where it's 1, and use those as the endpoints of the interval, since the rest of the graph will simply be an identical section to that one.

Sketch the given curves together y = 4^x, y^x, y=5^-x, y = (1/3)^x

x

Inverse Properties: tan(tan⁻¹(x)) = tan⁻¹(tan(θ)) =

x θ

Inverse Properties cos(cos⁻¹(x)) = cos⁻¹(cos(θ)) =

x θ

Inverse Properties: sin(sin⁻¹(x)) = sin⁻¹(sin(θ)) =

x θ

Inverse Trig Functions y = cos⁻¹x is equivalent to

x = cos y

Inverse Trig Functions y = sin⁻¹x is equivalent to

x = sin y

Inverse Trig Functions y = tan⁻¹x is equivalent to

x = tan y

Calculating the solutions for tanx

x = α + nπ

Unit circle definition cos θ =

x/1 = x

cot θ =

x/y

Reflections

y = -f(x) - reflection about y-axis y = f(-x) - reflect about x-axis

ln (y-9) - ln3 = x +lnx

y = 3x e^x + 9

Vertical and Horizontal scaling

y = c × f(x) : vertical stretch by c y = f(cx) : horizontal compression by c

Vertical and Horizontal shifts y = f(x) + c y = f(x+c)

y = f(x) + c (up by c) y = f(x+c) (left by c)

Which of the following functions are even or odd?

y = x² is even. We can see this graphically as it is symmetric about the y-axis. We can also see this algebraically, as (−x)² = x². • y = x³is odd. We can see this graphically as it is symmetric about the origin. We can also see this algebraically, as (−x)³ = −x³. • y = x + 1 is neither even nor odd. We can see this graphically as it is neither symmetric about the y-axis, nor the origin. We can also see this algebraically, as (−x) + 1 ≠ x + 1, and (−x) + 1 ≠ −(x + 1).

Unit circle definition sin θ =

y/1 = y

Unit circle definition tan θ =

y/x

Domain: The domain is all the values of q that can be plugged into the function cos θ =

θ can be any angle

Domain: The domain is all the values of q that can be plugged into the function sin θ =

θ can be any angle

Domain: The domain is all the values of q that can be plugged into the function sec θ =

θ ≠ (n + ½)π, n = 0, ±1, ±2

Domain: The domain is all the values of q that can be plugged into the function tan θ =

θ ≠ (n + ½)π, n = 0, ±1,±2

Domain: The domain is all the values of q that can be plugged into the function csc θ =

θ ≠ nπ, n = 0,±1,±2

Domain: The domain is all the values of q that can be plugged into the function cot θ =

θ ≠ nπ, n =0,±1,±2

180º

π (-1,0)

90º

π/2 (0,1)