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Click to start! Study Practice Quiz Application Welcome to beautiful… Function Junction

Click to start! StudyPracticeQuizApplication Welcome to beautiful…

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Page 1: Click to start! StudyPracticeQuizApplication Welcome to beautiful…

Click to start!

Study Practice QuizApplication

Welco

me to

beautif

ul…Function

Junction

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Hi! My name is Dale Derivative, and I’ll be working with you today! Before we head into the city, let’s

catch you up on what goes on around here!

Click to get into Dale’s Car

dx

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Let me introduce you to my car!

Click to go forward

Click to go back

Click to go home

dx Questions will Display Here

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Fish 0

411

The screens give solutions to my questions, just click them to select

an answer! Let’s calibrate:2+2=

dx

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X X

XX

You may want to try that Again, I’m not sure it worked…

dx

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O O

OO

Alright! I think we’re ready to head on out! Let’s see what there

is to know about this place! It’s okay if you didn’t read the

brochure.

dx

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Here in Function Junction, we’re crazy about functions! Not just how they work or look, though,

but how they MOVE

dx

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dx Just like the hills on the Function Junction roads, the functions have

a SLOPE, and it CHANGES every time you make the slightest movement on it; unless it’s

straight, of course.

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But you probably read all about that in the brochure! Let’s get into

the nitty-gritty…

dx

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Function Junction maps work a lot like your coordinate planes! Each

road’s a function, and here we study the CHANGE

dx

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dx

Just pull the screen out of the center console!

Before you Got here, I prepared a little in-depth look at how we

study the changes in our roads, and how you can help!

Screen

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x! xn

nxn+1f‘(x)

Great! So now that you know our purpose here, let’s see what you

can do!

What is the derivative

of f(x)?

dx

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X X

XX

Think About it this way, if f(x) is the original function, what do we

call the derivative?

dx

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O O

OO

Right! The Derivative of a function f(x) is f’(x)!

dx

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nxn+1 xn+1/n

nxn-1xn-1

Now for something a little more specific…

What is the derivative

of xn?

dx

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X X

XX

If you need a little help on this one, you can flip the screen back

up!

dx

Screen

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O O

OO

That’s right! You multiply x by the exponent and subtract 1 from the

exponent!

dx

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x5/4 4x3

x34x5

Alright, so let’s try something new…

What is the derivative

of x4?

dx

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X X

XX

Almost! You multiply by the exponent and subtract 1 from the

exponent!

dx

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O O

OO

That’s right! I think you’ve gotten the hang of it. Let’s move on to

WHY we use derivatives!

dx

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dx Just as a flat road may have a slope – uphill, downhill, or

straight – so, too, does a curved road!

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dx In Function Junction, most of our roads aren’t flat, and they move like functions. They usually look

like these!

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dx Here in Function Junction, our cars only move on the roads if

they know the slope of the road. Otherwise they could go right through! Lines are very thin!

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dx That’s why we, the people of Function Junction, need

derivatives!

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dx I have a boring mathematician’s explanation on the console screen so that you can better understand

how it all works!

Screen

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These tangent lines are how the car positions itself to move along functions.

dx

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(2,12) (2,4)

(2,16)(2,8)

So let’s guide you through creating a tangent line.

If f(x)=x3, and I’m at x=2, what is

my coordinate position?

dx

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X X

XX

Close. Think of it like this, if I’m on f(x), and I’m at x=2, what is my

position, (x,f(x))?

dx

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3x2 3x4

x2/3x4/3

Correct! Now, if I’m going to find the slope, I need the derivative!

What is f’(x) if f(x)=x3

dx

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X X

XX

We might need to change that answer around, or we’ll crash on

that function! The derivative of xn is nxn-1!

dx

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36 12

49

Right! So we know where we are, (2,8), and our derivative is 3x2,

so…

What is f’(2)

dx

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X X

XX

Try that again, our slope will be defined by f’(x)=3x2. To find f’(2),

plug in 2 for x!

dx

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y=12x y=12x-16

y=12x+16y=12x-8

Great! So one last step, to define the line!

dx Use the point-slope formula to define a line of

slope 12 that goes through (2,8)

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X X

XX

Close, the point-slope formula is (y-y1)=m(x-x1), so with our values,

simplify (y-8)=12(x-2)!

dx

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Correct! And that is our tangent line! Following these steps, you can get along any basic road in Function Junction! So now that we’re at the car rental store, go grab a car and get a move on!

Rent a car!

dx

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dx Alright! Now that you have your own car, you can get yourself around this town – if

you know how! I’ll meet you at my house, my wife, Iggy Integral and I have a spare room for you

while you vacation here!

Enter your new car!

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Your car is designed to drive itself. All you need to do as the pilot is navigate it by giving it a derivative and a tangent line for the point you

begin at on the curve. You have five curves ahead before you reach Dale’s place!

Your car will generate guesses of the approaching curves’ analytics.

You only have to select the correct guess.

Your car will deny incorrect guesses, and will remain stationary until it is certain of your answer.

All piloting will be controlled from this onboard navigation screen.

Good Luck!

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f’(x)= 2x

y=0

f’(x)= 2x3

y=0

f’(x)=2x3

y=1

f’(x)= x3/2

y=1

We are approaching a curve f(x)=x2. We will enter at the point (0,0). I need the derivative and the initial tangent line.

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ERRORUpon further analysis, the pilot’s judgment is found to be invalid.

The car will be stopped until the pilot’s judgment is corrected.

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f’(x)= x2/3

y=x/3-2/3

f’(x)= x4/3

y=x/3-2/3

f’(x)=3x2

y=3x+2

f’(x)= 3x4

y=3x+2

We are approaching a curve f(x)=x3. We will enter at the point (-1,-1). I need the derivative and the initial tangent line.

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ERRORUpon further analysis, the pilot’s judgment is found to be invalid.

The car will be stopped until the pilot’s judgment is corrected.

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f’(x)= 0

y=0

f’(x)= 1

y=x

f’(x)=x2

y=4x-6

f’(x)= x2

y=4x+6

We are approaching a curve f(x)=x. We will enter at the point (2,2). I need the derivative and the initial tangent line.

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ERRORUpon further analysis, the pilot’s judgment is found to be invalid.

The car will be stopped until the pilot’s judgment is corrected.

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f’(x)= 4x3

y=32x-56

f’(x)= x

y=2x+4

f’(x)=2x

y=4x

f’(x)= 4x

y=8x-8

We are approaching a curve f(x)=2x2. We will enter at the point (2,8). I need the derivative and the initial tangent line.

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ERRORUpon further analysis, the pilot’s judgment is found to be invalid.

The car will be stalled until the pilot’s judgment is corrected.

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f’(x)= 4x2/3

y=4x/3+8/3

f’(x)= 12x2

y=12x-8

f’(x)=12x4

y=12x+16

f’(x)= 4x2

y=4x

We are approaching a curve f(x)=4x3. We will enter at the point (1,4). I need the derivative and the initial tangent line.

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ERRORUpon further analysis, the pilot’s judgment is found to be invalid.

The car will be stopped until the pilot’s judgment is corrected.

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We have arrived at the destination point! You may turn off the car and enjoy your day!

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dx∫Welcome to our abode!

You’re welcome to spend the rest of your day doing whatever you want to do

around town! Just try to make it for

dinner, Iggy’s got something delicious

cooking!