Question

Graphs

Answer 1

2. In this problem, you will investigate the family of functions h(x) = x + cos(ax) , where a is a positive constant such that 0 < a < 4.

A. Graph the curves y(x) = x + cos(¼x) and y(x) = x + cos(4x).
B. For what values of a will h(x) have a relative maximum at x = 1?
C. For what value(s) of a will h(x) have an inflection point at x = 1?
D. Which values of a make h(x) strictly decreasing? Justify your answer.

Answer 2

I already drew the graphs.

Answer 3

Ok so where are we stuck? Part B?

Answer 4

yea @zepdrix

Answer 5

\[\Large\rm h(x)=x+\cos(ax)\]So for part B they're asking,
What a value gives us h'(1) = 0.

If our first derivative gives us a zero at x=1, that indicates that x=1 is a critical point, yes?
Oh they're going further as classifying it as a max, so maybe we'll have to a little bit more work beyond that.

Answer 6

\[\Large\rm h'(x)=1-a \sin(ax)\]Can you figure out this part?\[\Large\rm h'(1)=0, \qquad \text{solve for a}\]

Answer 7

Are we not using the equations in ine question a???

Answer 8

Correct, we are not.

In part a, those functions are y(x).
They're referring back to the general form h(x) in the description.

Answer 9

so if x = 1, then we just plug in 1 and then solve for a right?

Answer 10

Yah that's the idea I think..
It doesn't seem to be working out very nicely though hmm..

Answer 11

@zepdrix do we need to call for reinforcements? O:

Answer 12

Hmm probably :(
So like if you setup h'(1)=0 and `try to` solve for a, you get something like:\[\Large\rm 1=a \sin a\]yes?

Answer 13

I got the same... But it looks weird, no?

Answer 14

Yah I don't think there is any way to solve for a,
except by using some numerical approximation perhaps.

maybe I'm wrong .. hmmm what's going on here +_+

Answer 15

@Compassionate @AravindG @Hero @satellite73 Reinforcements : P lol

Answer 16

@theEric You said you took calculus?

Answer 17

\(\large h'(1)=1-a \sin(a)=1\)
Is what you guys have gotten too, which is great. Then you know that
\(\large a\sin(a)=0\) to make this true.
(If you don't see it intuitively, subtract \(1\) and then multiply by \(-1\))

There is two factors in this:
\(a\)
and
\(\sin(a)\)
and only one needs to be zero.
So, \(a=0\) or \(\sin(a)=0\).
What value(s) of \(a\) will make \(\sin(a)=0\)?

Answer 18

Hmm why'd you let h'(1) equal 1? :o

Answer 19

Oh, you're right. :) Misread!

Answer 20

\(\large h'(1)=1-a \sin(a)=0\)
\(1-a \sin(a)=0\)
\(\implies 1=a \sin(a)\)

Answer 21

Misread, but should've known! And this is tricky.

Answer 22

\(a\neq1,\quad a\neq0\)
But, hmm...
http://www.wolframalpha.com/input/?i=1%3Da+sina
scary stuff, but wolframalpha.com over complicates things sometimes.

Answer 23

Part (c) is easier. Maybe it would be best to move on for now? List what you have for partial credit, and go to (c)?

Answer 24

Hopefully someone else can help with (b) later.

Answer 25

Yes, we can move on for now.

Answer 26

What is true for inflection points?

Answer 27

I am not so sure myself lol, sorry. Can you elaborate?

Answer 28

Sure! Well derivatives find the rate of change, or slopes.
Keep that in mind.

The inflection points are the points at which the curve changes concavity.
Do you know what I mean by that?

Answer 29

Yes, like up or down.

Answer 30

Well, these would all be concave up, even though the middle segment is traveling down. So there is a distinction.

The slope is increasing is concave up, the slope decreasing is concave down. You got that? Look at those examples to check it out!