Function composition problem

[palmstierna]

Hello,

I badly need help with a question..

Just to start of by defining the functions.

f: Q -> R where f(x) = 0,2cos(pi*x)-7
g: N -> Q where g(x) = 5x/2

h(x) = f(g(x)) therefore h(x)= 0,2cos(5x*pi/2)-7

a) What is the range of h?
b) What is the domain of h?
c) Is h surjective (onto) ?

I would be very glad if anyone could help me out..

[Guest]

This is really just a matter of knowing the
"cosine" (of anything) can take all values between -1 and 1 so 0,2 cos()- 7 can take all values between -0,2- 7= -7,2 and 0,2- 7= -6.8. The range is -7.2 to -6.8, inclusive.

For x any positive integer (in N) we can multiply by 5/2 so we can take g of any positive integer. We can take the cosine of any number, multiply by 0,2 and subtract 7. Because the domain of g is given as N, the domain of h is N.

The domain was -7.2 to -6.8. That does not include all real numbers so, no, h is not surjective.

[Guest]

I don't believe that response is entirely correct. Yes, the range of "cos(x)" is -1 to 1 so the range of "0,2cos(5pi x/2)- 7" is 0,2(-1)- 7= -7,2 to 0,2(1)- 7= -6,8 where x ranges over all real numbers. But here, x must be a positive integer so the range may be restricted. We can look at specific values: when x= 1, y= 0,2cos(5pi/2)- 7= -7, when x= 2, y= 0,2 cos(5pi)- 7= -7,2, when x= 3, y= 0,2cos(15pi/2)- 7= -7, when x= 4, y= 0,2cos(10pi)- 7= -6,8, when x= 5, y= 0,2cos(25pi2)- 7= -7, etc. Get the idea? When x is odd, cos(5x/2) is 0 so y= -7. When x is a multiple of 4, cos(5x/2)= 1 so y= -6,8. When x is a multiple of 4 plus 2, cos(5x/2) is -1 so y= -7.2. The range is the set of those three numbers, {-7, -6.8, -7.2}.