Graph function of f(x) with max/min. Don't know what the problem is.

[SolCon]

Hi to all. :)

So I'm having a problem with this question, but in only one part, whereas the other parts of this question were done with correct answers. The only part where I'm having a problem, concerns drawing the graph of y=f(x). Here's the question:

Q. The function f is such that f(x) = a − b cos x for 0◦ ≤ x ≤ 360◦, where a and b are positive constants. The maximum value of f(x) is 10 and the minimum value is −2.

This came in parts (i) and (ii), both of which were done:

(i) Find the values of a and b.
Ans : a=4, b=-6

(ii) Solve the equation f(x) = 0.
Ans: x=131.8◦

(iii) Sketch the graph of y = f(x).
This is where I have a problem.

Here's how I tabulated the function:

http://usera.ImageCave.com/biosyn/grp%201.PNG

And here's the resulting graph:

http://usera.ImageCave.com/biosyn/grp%202.PNG

This seems to be wrong. However, I don't really know what the problem is and this is all I keep getting. :(

 

[tiny-tim]

Hi SolCon!

And here's the resulting graph:

Looks ok, except …

what happened to 3π/2 ? ​

 

[eumyang]

I can't see the attachments. Also,

Hi to all. :)
(i) Find the values of a and b.
Ans : a=4, b=-6

Why is b negative, when you stated before that a and b are positive constants?

 

[HallsofIvy]

The largest possible value for cos(x) is 1 so the smallest possible value for a- bcos(x) is a- b= -2. The smallest possible value for cos(X) is -1 so the largest possible value for a- bcos(x) is a+ b= 10.

Solve a- b= -2, a+ b= 10 for a and b.

 

[SolCon]

Apologies for the late reply.

Thanks everyone for the responses. :)

tiny-tim: Yes, I forgot about that. Was going by doubling the pi value. :)

eumyang: I don't know why you can't see the images, maybe something's blocking them at your end? Also, the part where the question asks for a positive b constant and me getting a -ve one is probably the part I'm having difficulty with.

HallsofIvy: I don't understand what you've done. Here's how I did it:

I agree with you about the fact that the greatest value of cos(x) is 1 [cos(0)]. So I did it like this:

> a-b cosx = f(x)
> a-b cos(0) = 10 [cos(0) = 10, i.e, greatest value]
> a-b(1) = 10
> a-b=10

The same with the least value of cos(x) which is -1 [cos(180)].

> a-b cos(180) = -2 [cos(180) = -2, i.e, lowest value]
> a-b(-1) = -2
> a+b = -2

This is how I got them and solved them simultaneously then. How come they not correct? :(

 

[HallsofIvy]

All I can say is that a= 4, b= -6 are the only values of a and b that will make a- bcos(x) have a maximum value of 10 and a minimum value of -2. There are not positive values that will work.

 

[HallsofIvy]

Apologies for the late reply.

Thanks everyone for the responses. :)

tiny-tim: Yes, I forgot about that. Was going by doubling the pi value. :)

eumyang: I don't know why you can't see the images, maybe something's blocking them at your end? Also, the part where the question asks for a positive b constant and me getting a -ve one is probably the part I'm having difficulty with.

HallsofIvy: I don't understand what you've done. Here's how I did it:

I agree with you about the fact that the greatest value of cos(x) is 1 [cos(0)]. So I did it like this:

> a-b cosx = f(x)
> a-b cos(0) = 10 [cos(0) = 10, i.e, greatest value]
> a-b(1) = 10
> a-b=10

The same with the least value of cos(x) which is -1 [cos(180)].

> a-b cos(180) = -2 [cos(180) = -2, i.e, lowest value]
> a-b(-1) = -2
> a+b = -2

This is how I got them and solved them simultaneously then. How come they not correct? :(

You did almost what I did except: subtracting a larger number gives you a smaller number! While 1 is the largest value of cos(x), it does NOT give the largest value of a- bcos(x), it gives the smallest.

A more formal way of seeing that is: $cos(x)\le 1$ for all x so, multiplying both sides of that by -b, $-bcos(x)\ge -b$ (remember that the inequality sign switches direction when you multiply both sides by a negative number) and then, adding a to both sides, $a- bcos(x)\ge a- b$. Since a- b is always less than or equal to a- bcos(x), it is the minimum value, not the maximum: a- b= -2, not 10.


Similarly, because $cos(x)\ge -1$, $-bcos(x)\le b$ and $a- bcos(x)\le a+ b$. Since a- bcos(x) is always less than or equal to a+ b, a+ b is its maximum value, 10.

Now solve a- b= -2, a+ b= 10.

 

[SolCon]

Once I again I thank you for the well structured explanation. It'll take me a while to completely allow the entire thing to penetrate through my skull, but I'm sure I'll get it.

Thanks for the help. :)