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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]

2 Answers

4 votes

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) (-∞,∞)

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 (-∞,∞)

User RaphaelS
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5.9k points
6 votes
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) ⇒⇒⇒⇒⇒ (-∞, ∞)










Match the functions with their ranges. Tiles y = 3sin(x − π) (-∞, ∞) y = 1 − sin(x-example-1
Match the functions with their ranges. Tiles y = 3sin(x − π) (-∞, ∞) y = 1 − sin(x-example-2
Match the functions with their ranges. Tiles y = 3sin(x − π) (-∞, ∞) y = 1 − sin(x-example-3
Match the functions with their ranges. Tiles y = 3sin(x − π) (-∞, ∞) y = 1 − sin(x-example-4
User Ibenjelloun
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5.9k points