Find a and b for f(x)=a-bcos(x), max is f(x)=10, min is f(x)=-2; find a, b

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Kayla65

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The function f is such that f(x)=a-bcosx for 0<= x<= 360, where a and b are positive constants.
The maximum value of f(x) is 10 and -2 at minimum.
Find a and b.

My workings
Solving for a:
10=a-bcosx
-a=-bcosx-10
a=bcosx+10

Solving for b:
-2=a-bcosx
0=a-bcosx+2
bcosx=a+2
[FONT=&quot]b=acos(x)^-1+2cos(x)^-1[/FONT][FONT=&quot].[/FONT][FONT=&quot]

[/FONT][FONT=&quot]Plug a into b:
b=(bcos(x)+10)cos(x)^-1+2cos(x)^-1
b=b+10cos(x)^-1+2cos(x)^-1
b=12cos(x)^-1+b
0=12cos(x)
( this is whe[/FONT]re I am stuck, assuming I am right with all my calculations)

If anyone could please help:?:
 
Kayla65 said:
The function f is such that f(x)=a-bcosx for 0<= x<= 360, where a and b are positive constants.
The maximum value of f(x) is 10 and -2 at minimum. Find a and b.
Click to expand...
The values of "a" and "b" are fixed. In particular, "a" is fixed. The value of -b*cos(x) varies -- between what two values? Then what must be the expression for the "max" value? What must be the expression for the "min" value?

These two expressions, each set equal to its given value, create two equations in two unknowns. Solve the linear system. ;)
 
stapel said:
The values of "a" and "b" are fixed. In particular, "a" is fixed. The value of -b*cos(x) varies -- between what two values? Then what must be the expression for the "max" value? What must be the expression for the "min" value?

These two expressions, each set equal to its given value, create two equations in two unknowns. Solve the linear system. ;)
Click to expand...

Are these 2 expressions
10=a-bcosx and -2=a-bcosx?

When you say -b*cos(x) varies between values are these + and -?
Im really stumped, sorry...
 
Kayla65 said:
The function f is such that f(x)=a-bcosx for 0<= x<= 360, where a and b are positive constants.
The maximum value of f(x) is 10 and -2 at minimum.
Find a and b.

My workings
Solving for a:
10=a-bcosx
-a=-bcosx-10
a=bcosx+10

Solving for b:
-2=a-bcosx
0=a-bcosx+2
bcosx=a+2
b=acos(x)^-1+2cos(x)^-1.

Plug a into b:
b=(bcos(x)+10)cos(x)^-1+2cos(x)^-1
b=b+10cos(x)^-1+2cos(x)^-1
b=12cos(x)^-1+b
0=12cos(x)
( this is where I am stuck, assuming I am right with all my calculations)

If anyone could please help:?:
Click to expand...
If you have 2 variables and one equation you can't solve for both variables. Think (yes think!) about the equation y = x+4. Well y = x+4 and x = y-4. While both are true statements you can't solve for both x and y. The reason you can't argue that now you have two equations (y = x+4 and x = y-4) so you can solve for x and y is because they are the same two equation--just written differently.

Let's for a moment assume that all your calculations are correct, so why are you stuck at 0=12cos(x). Is it because you don't know what to multiply 12 by to get zero or is it because you did not ask yourself 12 times what equals 0? Clearly 12*0 = 0 so cos(x)=0. To solve this just graph the cos(x) graph and see where it equals 0.

Now back to your problem. Somehow you need to use the two facts you were given (the max and min values) to solve for the two constants. I must be in a generous mood as I will give you a hint. What do a+b and a-b equal?. Please reply with these results and we will help you from there. Or even better, try to get further with this problem.
 
stapel said:
The values of "a" and "b" are fixed. In particular, "a" is fixed. The value of -b*cos(x) varies -- between what two values? Then what must be the expression for the "max" value? What must be the expression for the "min" value?

These two expressions, each set equal to its given value, create two equations in two unknowns. Solve the linear system. ;)
Click to expand...
I need to apologize to you. I am having a bad day and although I did read your post I did not realize it is exactly what I wrote below. It just look different. As always, you're always on top of things.
 
Is this correct?

Jomo said:
If you have 2 variables and one equation you can't solve for both variables. Think (yes think!) about the equation y = x+4. Well y = x+4 and x = y-4. While both are true statements you can't solve for both x and y. The reason you can't argue that now you have two equations (y = x+4 and x = y-4) so you can solve for x and y is because they are the same two equation--just written differently.

Let's for a moment assume that all your calculations are correct, so why are you stuck at 0=12cos(x). Is it because you don't know what to multiply 12 by to get zero or is it because you did not ask yourself 12 times what equals 0? Clearly 12*0 = 0 so cos(x)=0. To solve this just graph the cos(x) graph and see where it equals 0.

Now back to your problem. Somehow you need to use the two facts you were given (the max and min values) to solve for the two constants. I must be in a generous mood as I will give you a hint. What do a+b and a-b equal?. Please reply with these results and we will help you from there. Or even better, try to get further with this problem.
Click to expand...

Thank you so much for your help! :)
(I apologize if get this wrong)
So a+b=10 and a-b=-2
-2b=-12
b=6

a-6=-2
a=4
Am I correct? :mrgreen::confused:
 
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