Question

2. In this problem, you will investigate the family of function h(x)= x+cos(ax) where a is a positive constant such that 0 < a < 4
A. For what value of a will h(x) have a relative maximum at x=1?












B. For what value of a will h(x) have an inflection point at x=1?










C. Which values of a make h(x) strictly increasing? Justify your answer.


Another non-test but need to know how to solve question.

Answer 1

I will definately give medal for this question. I need to know how to solve these questions.

Answer 2

the problem is that i have no clue at all how to solve such a question.... never seen it before.

Answer 3

@amistre64 Would it be okay if you help me with one more question?

Answer 4

derivatives help; just treat "a" as a constant ... name it 5 if need be

Answer 5

or 3 since its nags us to be between 0 and 4 :)

Answer 6

so do the derivative of that formula? would the derivative be1-sin(ax)(a)?

Answer 7

yes

Answer 8

So then just plug in 1 and that would be the answer for 2. A right?

Answer 9

now we want to determine of there is any place where the slope is zero, or undefined for this derivative; this gives us "critical points" to test with

Answer 10

Okay... But I am not quite sure what I would need to be doing after this step.

Answer 11

1 - a sin(ax) = 0

a sin(ax) = 1

sin(ax) = 1/a

ax = arcsin(1/a)

x = (arcsin(1/a))/a

Answer 12

what is the largest "a" between 0 and 4 such that x=1?

Answer 13

woo.. tough question.

Answer 14

or rather what value of a produce and x=1

Answer 15

x = (arcsin(1/a))/a

1 = (arcsin(1/a))/a

a = arcsin(1/a)

sin(a) = 1/a

Answer 16

im thinking that maybe this function in "a" can be derived as well, but its an iffy thought

Answer 17

a cos(a) + sin(a) = 0

Answer 18

hum.... with the equation you gave me, I get this vague answer that say sin(a) = 1/a right? ...
Oh then sin(a)= 1/a so then cos(a) would basically mean -1/a right?

Answer 19

let me rethink this; since the value of a depends on x (x=1) to be exact

then we can consider a implicitly as a function of x
\[x+cos(ax)\]
\[1-sin(ax)(a'x+a)=0~:~x=1\]
\[1-sin(ax)(a'+a)=0\]
and do the algebra shuffle

Answer 20

\[1=sin(a)(a'+a)\]
\[csc(a)=a'+a\]
\[a'=csc(a)-a\]

Answer 21

which i believe is the same thing as sin(a) = 1/a

Answer 22

im at a loss on it at the moment ....

Answer 23

I can wait haha take your time. ty

Answer 24

i was thinking that the interval on x was from 0 to 4, we only need to determine a suitable value of a such that the slope of the line at x=1 is 0, relative to any epsilon beside it

Answer 25

what level is this?

Answer 26

h(x)= x+cos(ax) , assume a to be constant

h' = 1 - a sin(ax) = 0, at x=1

0 = 1 - a sin(a)

1/a = sin(a)

i get 2 as by graphing that fit, one is sure to be a min and the other a max

Answer 27

x = about 2.7726

http://www.wolframalpha.com/input/?i=y%3Dx%2Bcos%28%282.7726%29x%29%2C+x%3D0+to+2

x = about 1.11416

http://www.wolframalpha.com/input/?i=y%3Dx%2Bcos%28%281.11416%29x%29%2C+x%3D0+to+2

Answer 28

haha this is just calculus ab class.

Answer 29

so do we do the second derivative test now for this?

Answer 30

I think 1.114 would be the max because the original curve has highest point here?

Answer 31

yeah, we can try that, it might actually help us out ... might not :)

h' = 1 - a sin(ax)

h'' = a^2 cos(ax)

Answer 32

yes, but how to determine an exact value for a instead of an approximation ....

Answer 33

h'' = -a^2 cos(ax) that is

Answer 34

I really appreciate your effort to really help me out.

Answer 35

at x=1 thats

h'' = -a^2 cos(a)

we would need the property such that any value a little bit to the left and right of a is ?

Answer 36

youre welcome so far :)

Answer 37

yes. I think that when the left side is sloping upward and right side is sloping downward that would be the max?

Answer 38

slope relates to h'

how does concavity help us? the rate of change of the slope

Answer 39

i always have to refer to y=x^2; since i know its cave up, and min

y = x^2
y' = 2x
y'' = 2, is positive ; so when y'' is negative, we have cave down and max

Answer 40

since -a^2 is always negative, we would want to confine cos(a) such that:

0 <= a <= pi/2

Answer 41

woo.... quite tough..... So let's put aside this question for now. So the next question ask what value of a will h(x) have an inflection point at x=1....... So this would be something related to previous question right?

Answer 42

h'' = 0, or undefined are critical points for inflection

Answer 43

THen as you have said, C asks which values of a makes h(x) strictly increasing. So that would mean the things that we have been previously talking about right?

Answer 44

that would tie in to A and B yes

Answer 45

h' >= 0 is always increasing

Answer 46

So for the inflection question, we need to find the value of a that makes h''=-a^2cos(ax) 0 right? then 0 would be the answer ?

Answer 47

h'' = 0 for inflection @ x=1 yes

Answer 48

brb

Answer 49

recall that a is not 0 and not 4, but between 0 and 4, unless you misposted that information

Answer 50

oh right... That means I would need to be finding another value that fits this...

Answer 51

ill have to think about this and come back to it, most likely develop an idea tonight and post it tomorrow

Answer 52

okay thanks I will be waiting

Answer 53

@amistre64 Still no clue?

Answer 54

nothing elementary :)

i considered csc(a) = a, which is the intesection of y = csc(a) and y=a

since y=a has a slope of 1, i thought; csc'(a) = 1 and came up with the golden ratio:

x = (1 +- sqrt5)/2

so i thought maybe by finding that point i could interpolate for csc(a) = a