Calculus Chapter 2 Study Guide

If a rock is thrown upward on the planet Mars with a velocity of 10m/s, its height in meters t seconds later is given by y=10t-1.86t²; find the average velocity over the given time intervals: [1,2], [1,1.5]

(10(2)-1.86(2)²) - (10(1)-1.86(1)²)/2-1 = 4.42; (10(1.5)-1.86(1.5)²)-(10(1)-1.86(1)²)/1.5-1 = 5.35

limh->0 ((2+h)³-8)/h

0/0 when plug in 0; (2+h)(2+h)(2+h); simplify down and get limh->0 (h²+6h+12)=12

limx->4 (x²-4x)/(x²-3x-4)

0/0 when plug in 4; x(x-4)/(x-4)(x+4); limx->4 x/x+1 = 4/5

limx->0 (1/t - 1/t²+t)

0/0, do more, multiple 1/t by t+1/t+1; get 1

limx->2 √4u+1 - 3/u-2

0/0, do more, multiply bu √4u+1 + 3; get 2/3

limx->1 t⁴-1/t³-1

0/0, do more; different of cubes (SOAP); (t-1)(t²+1+1); get 2

limx->0 (√x+4 - 2/x)

0/0, do more; limx->0 (√x+4 -2/x)(√x+4 -2/x)/(√x+4 -2/x); limx->0 1/(√x+4 - 2) = 1/4

limx->-4 √x²+9 - 5/x+4

0/0, multiply by √x²+9 + 5/√x²+9 + 5; get -8/√5 + 5

limh->0 (3+h)⁻¹ - 3⁻¹/h

0/0; 1/(3+h)(-3)+h; -1/9

Limx->0 (sin3x/x) =

1

Explain why discontinuous at the given number a and graph: {x^2 -x/x^2-1 if x does not equal 1 {1 if x-1

1) f(1)=1; 2) limx->1 f(x) = 1/2; 3) limx->1f(x) does not equal f(1); therefore, f(x) is discontinuous at x=1

Use intermediate value theorem to show that there is a root of the given equation in the specified interval: x^4+x-3=0, (1,2)

1) f(x) is a polynomial which is continuous on (-infinity, +infinity); 2) f(1) = -1<0, f(2) =15>0; therefore, there is a c on the interval (1,2) such that f(c)=0

limx->0⁺ (1/x-1/|x|)

1/x - 1/x x>0 1/x - 1/-x x<0 0, x>0 2/x, x<0; lim=0

f(x) = {x+2, x≥3 {-x+7, x<3 limx->1f(x) =

6; if at 3, check both sides

If g is continuous at a and f is continuous at g(a), then the composite function f of g given by (f of g)(x)= f(g(x)) is continuous at

A

When are functions continuous

A function is continuous at a number a if limx->a f(x) = f(a)

Show that limx->0 |x| = 0

Always split absolute value; f(x) = |x| = {x, x≥0 - right {-x, x<0 - left limx->0⁻ f(x) = -0=0 limx->0⁺ f(x) = 0

Direct substitution property

Any polynomial is continuous everywhere (-infinity, +infinity); any rational function is continuous wherever it is defined; that is, it is continuous on its domain

Limit

Approaching from both sides

Where is the function discontinuous: f(x) = [[x]]

At all integers, removable; {x|xEZ

Where is the function discontinuous: f(x) = {x²-x-2 if x≠2 {1 if x=2

At x-2, removable; actual point and then limit defined at open interval

Where is the function discontinuous: f(x) = {1/x² if x≠0 {1 if x=0

At x=0, infinite; do the three steps, limit exists and f(a) exists but don't equal each other

Where is the function discontinuous and what type: f(x) = x²-x-2/x-2

At x=2, removable; hole at 2; 3 steps

Why is it called a removable function?

Because the discontinuity cane removed by redefining f at a single number (where the hole is)

limf(x)x->2

Both sides must equal same number to get both

limx->-2 2-|x|/2+x

Check from right and left; 2-x/2+x x≥0, (2+x)/(2+x) x<0; get 1; check both sides because constant/0

Explain why discontinuous at a and sketch graph. F(x) = {x+2 if x<0 {e^x if 0 is less than or = to x less than or = to q {2-x if x>1

Discontinuous at x=0, x=1 overall because jumps; continuous at x=1 from right, x=0 from right, x=-2 from left

If f is defined near a (defined on an open interval containing a), we say that f is______at a if f is not_____at a

Discontinuous; continuous

0/0

Do more

3 steps to see if continuous

F(a) is defined (a is in the domain of f); limx->a f(x) exists; limx->a f(x) = f(a)

If f is continuous at b and limx->a g(x)=b, then limx->af(g(x)) =

F(b); limx->af(g(x))=f(limx->ag(x))

If f and g are continuous at a and c is a constant, the what else is continuous at a

F+g; f-g; cf; fog; f/g (if g(a) does not equal 0)

limf(x)x->2⁻

From the left

limf(x)x->2⁺

From the right

Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450m above the ground. Find the velocity of the ball after 5 seconds

Galileo's law (s(t)=4.9t²); t=5, single instant of time, no time interval, so find points closest to tangent line, approximate, use t=5.1; m=4.9(5.1)²-4.9(5)²/.01 = 49.49m/s; do that for a couple of different numbers (5.1, 5.05, 5.01, 5.001); about 40 - likely instantaneous velocity

Continuity in geometry

Graph with no break in it; draw without removing pen from paper

Find limx->1 g(x) where: g(x) = {x+1 if x≠1 {π if x=1

Hole at (1,2); point at (1,π); graph; limit = 2 (see from graph)

Estimate the value of limx->0 (√t²+9 - 3/t²)

Hole at 0; plug in 0, get 0/0, do more; limt->0 (√t²+9 - 3/t²)(√t²+9 +3)/(√t²+9 + 3); get 1/6 when simplify and plug in 0; limt->0 (1/√t²+9 + 3) = 1/6

Derivative

How to find the slope of that one point, tangent line

Direct substitution property of limits

If f is a polynomial or a rational function and a is in the domain of f, then limx->a f(x) = f(a)

A function is continuous on an interval if

It is continuous at every number in the interval

3 types of discontinuity

Jump discontinuity: point and then undefined point at same x value, limit is DNE; removable discontinuity: there is a single point and function with undefined point, limit exists, f(a) exists but limx->a f(x) ≠ f(a); infinite discontinuity: there is a point and infinity, limit exists at some infinity, f(a) exists at actual value, limx-a f(x) ≠ f(a)

Power law of limits

Limx->a [f(x)]ⁿ = [limx->a f(x)]ⁿ

4 times when to check both sides

Limx->a f(x) on a graph; limx->a f(x) when your substitute gives you a constant over 0; piecewise functions; absolute value

limt->0 (1/t√1+t - 1/t)

Multiply 1/t by √1+t/√1+t; simplify and get -1/2

limx->16 4-√x/16x-x²

Multiply by 4+√x/4+√x; simply and get 1/128

Find an equation of the tangent line to the parabola y=x² at the point P(1,1)

Need 2nd point; (x,x²) (y=x²); m=x²-1/x-1; m=2, y-1=2(x+1)

Does limx->3 [[x]] exist?

No; limits for integers for greatest integer function don't work, but in between integers yes; number line: 2, 3, 4; for left, always round down to 2, and for right, always round down to 3 so doesn't exist

Is the graph of tangent continuous?

No; vertical asymptotes; infinite discontinuities at +- pi/2, +- 3pi/2, +- 5pi/2

limx->-1 x²-4x/x²-2x-4

Plug in -1 and get undefined, check both sides; VA at x=-1, check both sides; number line (-2, -1, 0); plug in -2 for form left and get ∞, plug in 0 for from right and get -∞; DNE

Evaluate: limx->-2 (x³+2x²-1/5-3x)

Plug in -2; in domain, so don't have to do more; limx->-2 (x³+2x²-1/5-3x) = -1/11

Find limx->1 (x²-1/x-1)

Plug in 0, get 0/0, do more; limx->1 (x+1)(x-1)/(x-1); limx->1 (x+1) = 2

Guess the value of limx->0 (sinx/x)

Plug in 0, get 0/0, do more; squeeze theorem, limx->0(sinx/x)=1; limθ->0 (sinθ/θ) = 1

limx->-1 2x³+3x+1/x²-2x-3

Plug in 01 and get 0/0; do more and get 1/4

Evaluate: limx->5 (2x²-3x+4)

Plug in 5; 2(5)²-3(5)+4; limx->5 (2x²-3x+4) = 39

How to make the function continuous: f(x) = {x²-x-2 if x≠2 {1 if x=2

Removable discontinuity; f(2) = 3; make y = 3 (or something), fill in the hole

Sketch graph of function that satisfies all conditions: limx->0 f(x)=1, limx->3⁻f(x)=2, limx->3+f(x)=2, f(0)=-1, f(3)=1

See 2.2 homework; just graph what it says to

Are trigonometric functions continuous at every number in their domain?

Sin, cosine but not tangent

How define f(2) in order to make f continuous: f(x) = x^3-8/x^2-4

Solve and get 3 after simplify; f(x) = {x^3-8/x^2-4 if x does not equal 2 {3 if x=2 Fill in the hole (removable discontinuity) at f(2) = 3

limx->3 (2x+|x-3|)

Split; f(x) = 2x + |x-3| = {2x+x-3, x-3≥0 {2x+ -(x-3), x-3<0 Simply and get 6

Prove that limx->0 |x|/x does not exist

Split; {x/x, x>0 {-x/x, x<0 limx->0⁻ f(x) = -1; limx->0⁺ f(x) = 1; DNE

Intermediate value theorem = verbatim and know graph

Suppose that f is continuous on the closed interval [a,b] and let N be any number between f(a) and f(b), where f(a) does not equal f(b). Then there exists a number c in (a,b) such that f(c) = N

A continuous process is one that

Takes place gradually, without interruption or abrupt change

Why is it called a jump discontinuity?

The function jumps from one value to another

Constant multiple law of limits

The limit of a constant times a function is the constant times the limit of the function

Difference law of limits

The limit of a difference is the difference of the limits

Product law of limits

The limit of a product is the product of the limits

Quotient law of limits

The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0)

Sum law of limits

The limit of a sum is the sum of the limits

limf(x)x->a = L

The limit of f(x) as x approaches a equals L

Limx->0 (sin9x/x)

Times 9/9; get sin9x/9x = 9(1)=9

Rule about limits and plugging in

When can plug in and fits in domain, plug in

Are polynomials continuous at every number in their domain?

Yes

Is the graph of cosine continuous?

Yes

Is the graph of sine continuous?

Yes

Are rational functions continuous at every number in their domain?

Yes except where domain = 0

Are root functions continuous at every number in their domain?

Yes unless square root of a negative number

Are physical phenomena usually continuous? Example

Yes; displacement or velocity of a vehicle varies continuously with time as does a person's height; but discontinuities occur with electric currents

Can there be another point at the same value of a

Yes; hole in graph at actual L; not what = to, so disregard with limit

Intermediate value theorem - can f(c) = 0, what values of a and b for this

Yes; y value of a must be negative and y value of b positive

limx x->a =

a

limx->a xⁿ =

aⁿ

limx->a c =

c

limx->a [cf(x)] =

climx->a f(x)

limx->0.5⁻ 2x-1/|2x³-x²|

f(x) = {2x-1/x²(2x-1), 2x-1>0 {2x-1/-x²(2x+1), 2x-1<0 Simplify and get limx->0.5⁻ 1/x² = -4

limx->-2 [f(x)+5g(x)]

limx->-2 f(x) + 5limx->-2 g(x)

lim(x-1/x²-1)x->1

limx->1 (x-1/(x+1)(x-1)); hole at 1, fine with limits; limx->1 (1/x+1) = 1/2; plug in 1, undefined 0/0 -> do something else; (1,1/2); simplify further

Find the limit of g(x) when x-> 1 g(x) = {x-1/x²-1 if x≠1 {2 if x=1

limx->1g(x)=1/2; g(1)=2; make graph

How to find limit of f(x)=x²-x+2

limx->2 f(x) = 4; look at table of values and see what y approaches from both sides of 2

f(x) = {√x-4 if x>4 {8-2x if x<4 Determine whether limx->4 f(x) exists

limx->4⁻ f(x) = 0 (8-2(4) =0) Limx->4⁺ f(x)=0 (√4-4 = 0) limx->4 f(x) = 0

limx->a [f(x) + g(x)] =

limx->a f(x) + limx->a g(x)

limx->a [f(x) - g(x)] =

limx->a f(x) - limx->a g(x)

limx->a [f(x)g(x)] =

limx->a f(x) × limx->a g(x)

limx->a f(x)/g(x) =

limx->a f(x)/limx->a g(x) if limx->ag(x) ≠0

Root law of limits

limx->a ⁿ√f(x) = ⁿ√limx->a f(x)

limx->0 tan3x/tan5x

limx->af(x)/limx->ag(x); limx->0 sin3x/cos3x / sin5x/cos5x; limx->0 sin3x/limx->0cos3x / limx->0 sin5x/limx->0cos5x; plug in 0, cos of 0=1, so cos cancel out; limx->0 sin3x x 3x/3x / limx->0sin5x x 5x/5x; limx->0 3xsin3x/3x / limx->0 5xsin5x/5x; 3(1)/5(1) = 3/5

A function f is continuous from the right at a number a if:

limx->a⁺ f(x) = f(a)

A function f is continuous from the left at a if

limx->a⁻ f(x) = f(a)

limh->0 1/(x+h)² - 1/x²/h

x²-(x+h)²/(x+h)²x² x 1/h; -2/x³

limx->a ⁿ√x =

ⁿ√a