Question

More calculus....(posting problem below)

Edit: I just need help with C and D now.

Answer 1

I've got the graph, but I'm struggling to understand what exactly they're asking for in (b).

Answer 2

@math&ing001
I just wanted to give you a heads up that Openstudy doesn't like me, so every now and then my connection will cut out. Because of this, sometimes it takes me a little longer to respond.

Answer 3

It's alright ^^

Answer 4

They are asking for the distance between f and g, and since it's variable they want you to express it in function of x. If you post the graph, I can respond on it.

Answer 5

I can't get the domains to show up, but here are the functions graphed...

Answer 6

It's as easy as writing for \(x \in [0,4]\) :
d(x) = f(x) - g(x)

Answer 7

So I would subtract the second function from the first, and that would be my distance?

Answer 8

Yep !

Answer 9

\[(\frac{1}{2}x^2)-(\frac{1}{6}x^4-\frac{1}{2}x^2)= distance\]
\[(-\frac{1}{6}x^4+x^2)\]

Answer 10

It's 1/16 though

Answer 11

How did you get that if I might ask?

Answer 12

Basic math Vertical distance between A and B is d=y1-y2

Answer 13

How would you know what y values to subtract?

Answer 14

Distance is always positive ! To make sure you get it right, it's preferable to use d = |y1-y2|.

Answer 15

I did y1-y2 without the absolute value because I saw in the graph that f(x) is always superior or equal to g(x) between 0 and 4.

Answer 16

I'm sorry, but I'm confused as to what you're saying.

Answer 17

I've gone back through what you said, but I still don't understand where you got 1/16 from.

Answer 18

@Catseyeglint911 1/16 is on your sheet. g(x) = 1/16 x^4 - 1/2 x^2

Answer 19

Oh no! For some reason I was reading it as 1/6, and it confused me.

Answer 20

Just to be sure...How did you get the x^4 off the 1/16?

Answer 21

Wait...Just by looking at the graph, it seems that the maximum distance between the 2 graphs is greater than 1/16.

Answer 22

Where you assuming that 1/16 is the max ?

Answer 23

Were*

Answer 24

I guess I was. I have a funny feeling I'm wrong though.

Answer 25

Well they're asking you to find it by calculus, so we need to find the derivative of d(x), then find critical points.

Answer 26

Correct me if I'm wrong (I probably am)
First, you subtract g(x) from f(x), and the answer will be d(x). Next, you find the derivative of d(x) and find the critical points?

Answer 27

Yes ! Don't forget that the critical point should be between 0 and 4 ;)

Answer 28

Oh okay. So would it be... \[−\frac{1}{6}x^4+x^2\]?

Answer 29

And I would take the derivative of that?

Answer 30

Again with the 1/6 man !
\[d(x) = -\frac{ 1 }{ 16 }x ^{4} + x ^{2}\]

Answer 31

Grrrr! I can't believe I did that again! Sorry.
I have it written down correctly. I just can't seem to type it right.

Answer 32

I got \[\frac{1}{4}x^3\] as the derivative, and I got an critical point of x=0.

Answer 33

Is that correct?

Answer 34

Not quite, you forgot the derivative of x^2

Answer 35

...and the minus sign

Answer 36

Thanks for pointing that out. I fixed the derivative, and I'll go solve for the critical point

Answer 37

Awesome :)

Answer 38

Okay. I'm having some troubles solving for the critical point. I know to set the denominator. So far I have \[x=(4)\sqrt[3]{-2x}\] How would I simplify it though?

Answer 39

What did you get for the derivative ?

Answer 40

I got \[\frac{1}{4}x^3+2x\]

Answer 41

Yeah, just a missing minus sign. \[-\frac{ 1 }{ 4 }x ^{3}+2x = 0\]
Have you tried factoring by x ?

Answer 42

\[-x\left(\frac{x}{2}+\sqrt{2}\right)\left(\frac{x}{2}-\sqrt{2}\right)\]

Answer 43

Is there anything else I can do to it?

Answer 44

Yeah, that's correct ! Now put that equal to zero and solve for x =)

Answer 45

Common, you're really close !

Answer 46

Okay I got them

Answer 47

Okay I got them\[x=0, x=-2\sqrt{2}, x=2\sqrt{2}\]

Answer 48

Yep ! Now remember \(x _{\max}\) should be between 0 and 4.

Answer 49

\[x=2\sqrt{2}\]

Answer 50

You got it !

Answer 51

@jim_thompson5910
Could you help me with C and D by chance?

Answer 52

sure, let me read it over

Answer 53

Thank you!

Answer 54

if x = 2*sqrt(2), then what is f ' (x) equal to?

Answer 55

\[2(2√2)\] = \[4√2\]

Answer 56

Should I solve for g'(x) as well/

Answer 57

f(x) is x^2 over 2

f ' (x) = 2x/2 = x

Answer 58

So just 2√2?

Answer 59

yes, now do the same for g(x)

Answer 60

\[\frac{1}{4}(2\sqrt{2})^3-(2\sqrt{2})\]

Answer 61

what do you get when you simplify that?

Answer 62

Just a second

Answer 63

\[2\sqrt{2}\]

Answer 64

yep so this means that the two tangent lines are parallel

Answer 65

This is not a coincidence

Conjecture: the tangent lines on f(x) and g(x) are parallel at the x value where the distance from f(x) to g(x) has been maxed out.


Proof:

Let d(x) = f(x) - g(x) be the distance between the two functions. Assume we're restricted on an interval where f(x) > g(x) for every x in this interval.



Apply the derivative
d(x) = f(x) - g(x)
d/dx[d(x)] = d/dx[f(x) - g(x)]
d/dx[d(x)] = d/dx[f(x)] - d/dx[g(x)]
d '(x) = f '(x) - g '(x)


the distance d(x) is maxed out when d '(x) is zero, so

d '(x) = 0
f '(x) - g '(x) = 0
f '(x) = g '(x)


telling us that when the distance is maxed out, the derivatives of f and g are equal. The derivative tells us info about the slope of the tangent line. So if the derivatives are equal at that critical x value, then the slopes of the tangent lines are equal at that critical x value.

Answer 66

So the slope of the tangent lines would be\[2\sqrt{2}\]

Answer 67

yes

Answer 68

I would plug those slopes into the formula y-y1 = m(x -x1) correct?

Answer 69

correct

Answer 70

you'll need to know (x1,y1)

x1 is the critical x value
y1 is the output after plugging the critical x value back into the original function (f or g)

Answer 71

So the x values will also be \[2\sqrt{2}\]

Answer 72

yes

Answer 73

I got 4 for f(x)

Answer 74

so what is the equation of the tangent line for f(x)

Answer 75

\[y = 2\sqrt{2}x - 4\]

Answer 76

I go this for g(x)...\[y = 2\sqrt{2}x - 8\]

Answer 77

Now what you said earlier about the conjecture, does that deal with D?

Answer 78

that is correct