Finding Domain and Range of Functions

[MorganJ]

Find the domain and range of y=sin^2x


I have a hard type computing this into my graphing calculator. Can someone help in the steps to find the domain?

 

[rock.freak667]

y=sin²x=(sinx)².

since it is squared it is never negative. The domain should be easy to find, given your knowledge of y=sinx

 

[MorganJ]

So the domain would be 0 less than or equal to positive infinity? and the range would be the same?

 

[rock.freak667]

So the domain would be 0 less than or equal to positive infinity? and the range would be the same?

What is the domain of sin(x)? The range is the values of 'y' that the graph lies between.

You know that -1≤ sinx ≤ 1, so where would sin²x lie between?

 

[MorganJ]

the domain of sin (x) is -∞ < x < ∞ and would sin²x lie between -2 ≤ x ≤ 2...sorry i am bit confused.

 

[rock.freak667]

the domain of sin (x) is -∞ < x < ∞

Right, sin²x has the same domain as sin(x)


and would sin²x lie between -2 ≤ x ≤ 2...sorry i am bit confused.[/QUOTE]


if y=sin²x, what is it's maximum and minimum value?

 

[MorganJ]

Wouldn't its maximum value be 1 and the minimum value be -1?

 

[rock.freak667]

Wouldn't its maximum value be 1

yes

and the minimum value be -1?

remember, sin²x = (sinx)² so it is a perfect square, meaning sin²x ≥ 0 . So what's the minimum value going to be?

 

[MorganJ]

So...would it be 1 as well?

 

[rock.freak667]

So...would it be 1 as well?

no

in y=x², what is the lowest value of y which gives a real value for x?

 

[MorganJ]

I have no clue, honestly. I am graphing it on my calculator. Would it be zero? I do not think the minimum value is a negative number.

 

[rock.freak667]

I have no clue, honestly. I am graphing it on my calculator. Would it be zero? I do not think the minimum value is a negative number.

yes it would be zero.

 

[MorganJ]

Okay so the domain would be -∞ < x < ∞ and the range is 0 ≤ y ≤ 1 ?

 

[rock.freak667]

Okay so the domain would be -∞ < x < ∞ and the range is 0 ≤ y ≤ 1 ?

That should be correct. You could write $x \epsilon \Re$ as well I believe.

 

[MorganJ]

Okay. Thank you so much for helping, rock.freak667. I really appreciate it :-)