Question

Match the functions with their ranges. Tiles y = 3sin(x − π) (-∞, ∞) y = 1 − sin(x) [-1, 7] y = 3 + 4cos(x − π) [-3, 3] y = 2 + cot(x) [0, 2]

Answer 1

Answer:

    Function                     Range

 y = 3sin(x − π)                 [-3,3]

y = 1 − sin(x)                     [0,2]

 y = 3+4cos(x − π)             [-1,7]

 y = 2 + cot(x)                    (-∞,∞)

Step-by-step explanation:

We know that the range of:

y= sin x and y=cos x is : [-1,1]

and the range of the cot function is : (-∞,∞)

1)

y = 3sin(x − π)

As we know that:

  -1≤  sin x ≤1

⇒    -1 ≤ sin(x-π) ≤1

⇒     -3 ≤ 3sin(x-π) ≤3

Hence, Range is [-3,3]

2)

   y = 1 − sin(x)

     -1≤  sin x ≤1

⇒    1 ≥ -sin x ≥ -1

( since on multiplying by -1 the inequality gets reversed in sign)

It could be written as:

 -1 ≤ -sin x ≤ 1

⇒   1-1 ≤ 1-sin x ≤ 1+1

⇒   0 ≤ 1-sin x ≤ 2

 Hence, Range=[0,2]

3)

y = 3 + 4cos(x − π)

  As we know that:

         -1 ≤ cos x ≤ 1

          -1 ≤ cos (x-π) ≤ 1

so,       -4 ≤ 4 cos(x-π) ≤ 4

so,  3-4≤ 3+4cos(x-π) ≤ 4+3

⇒     -1≤ 3+4cos(x-π) ≤ 7

Hence, Range= [-1,7]

4)

       y = 2 + cot(x)

as we know that:

               cot x lie between (-∞,∞)

so adding a constant won't change it's range.

It remains same (-∞,∞)

Answer 2

By graphing each function, we can find the range of each one
So,
     From figure 1:  The range of  y = 3sin(x − π)         ⇒⇒⇒⇒⇒ [-3, 3] 
     From figure 2:  The range of  y = 1 − sin(x)           ⇒⇒⇒⇒⇒ [0, 2]
     From figure 3:  The range of  y = 3 + 4cos(x − π)  ⇒⇒⇒⇒⇒ [-1, 7]
     From figure 4:  The range of  y = 2 + cot(x)           ⇒⇒⇒⇒⇒ (-∞, ∞)