What is the domain and range ofy=1/xsquare

Ad

{x|x is element of C, x not= 0}

Posted on Jun 15, 2009

Ad

Give the domain and range of this relation

{(0,2),(4,5),(2,9)}

Posted on Jan 29, 2009

Ad

Hi,

a 6ya Technician can help you resolve that issue over the phone in a minute or two.

Best thing about this new service is that you are never placed on hold and get to talk to real repair professionals here in the US.

click here to Talk to a Technician (only for users in the US for now) and get all the help you need.

Goodluck!

Posted on Jan 02, 2017

The arccosine function is defined for arguments in the range from -1 to +1. 50/30 is greater than 1 and thus is out of the domain.

Feb 06, 2014 | Texas Instruments TI 30XIIS Scientific...

The arccosine function is defined for arguments in the range -1 to +1. -125.85/-120.36 is larger than 1 and hence outside the domain of the arccosine function.

Dec 03, 2013 | Texas Instruments TI-84 Plus Calculator

The sine and cosine function have a range between [-1, 1]. The domain of their inverse functions is [-1,1]. So 20/1 which is 20 is out of the domain of definition of the functions. No limitations for tangent and cotangent.

Oct 22, 2013 | Casio FX-115ES Scientific Calculator

Is there a question? Why so secretive, spill it out.

To access the arc cosine (cos^-1) press 2nd COS and enter the argument. Make sure you are within range, because the domain of arc cosine is from -1 to +1. Any value outside of this closed interval will warrant a DOMAIN ERROR.

To access the arc cosine (cos^-1) press 2nd COS and enter the argument. Make sure you are within range, because the domain of arc cosine is from -1 to +1. Any value outside of this closed interval will warrant a DOMAIN ERROR.

May 29, 2011 | Texas Instruments TI-83 Plus Calculator

Since you are familiar with sines, cosines, you know that their ranges (interval of values) varies from -1 to 1. The inverse functions of sine and cosine tkae their values in that very domain, [-1,1].

However you fed the arc sine function (sin^-1) a vlaue of (25/20.48) and that value is obviously larger outside the [-1,1] domain, hence the DOMAIN error message.

No such domain limitations exist for arc tangent (tan^-1) because the range of the tangent function spans the open interval ]negative infinity to positive infinity[.

However you fed the arc sine function (sin^-1) a vlaue of (25/20.48) and that value is obviously larger outside the [-1,1] domain, hence the DOMAIN error message.

No such domain limitations exist for arc tangent (tan^-1) because the range of the tangent function spans the open interval ]negative infinity to positive infinity[.

Nov 02, 2010 | Texas Instruments TI-84 Plus Silver...

The cosine function takes its value on the whole real line, the angle domain spans ]-infinity, + infinity[. The range of the function is however limited to the [-1, +1] interval.

Thus if you want to calculate the arc (angle) from the value of the cosine, the cosine must be in the interval [-1, +1]. Any value outside this domain will give you a domain error.

Check that the values you feed to the arc cosine function are in the interval [-1,+1].

If the angle unit is the degree the value returned bay [cos^-1] is in degrees, and if the angle unit is the radian, the value returned is in radians

Thus if you want to calculate the arc (angle) from the value of the cosine, the cosine must be in the interval [-1, +1]. Any value outside this domain will give you a domain error.

Check that the values you feed to the arc cosine function are in the interval [-1,+1].

If the angle unit is the degree the value returned bay [cos^-1] is in degrees, and if the angle unit is the radian, the value returned is in radians

Mar 04, 2010 | Texas Instruments TI-89 Calculator

Feb 26, 2010 - I was right to suggest to
you to read the page on domain and range of functions: it would have
clarified the concepts to you.

The domain of the sine function is from -infinity to + infinity. But since the function is periodic, with a period equal to 2Pi, by limiting the DOMAIN of values to -1*Pi to +1*PI you see all there is to see. All the rest can be obtained by translation of the curve.

The RANGE of the sine function is LIMITED to values in the interval [-1, 1]

Let us summarize: The DOMAIN of the sine function is ]-infinity, +infinity[ and its RANGE is [-1,+1].

That being said, there is something I would like to point to you

These are the numbers.

You want a "square", so be it. Here is the window setting

and the corresponding picture. Does it look like a square?

Why do you insitst on drawing a square? Horizontally you have the angle ( a number with a unit), while vertically you have a ratio of two lengths ( a pure number). Would even think about a square if you drew your sine function with the degree as angle unit. Horizontally you would have a domain [-180 degrees, 180 degrees] while vertically you have a range [-3.14..., +3.14...]. How can that be a square?

I showed you how you can fix every dimension in the graph window (see the first picture) . Choose any values that you believe will give you a square graph. And I do mean to say "that make you believe", because there is no meaning attached to the "fact" that the window looks like a square. An angle cannot be compared to a the projection of one side of a right triangle onto the hypotenuse.

The domain of the sine function is from -infinity to + infinity. But since the function is periodic, with a period equal to 2Pi, by limiting the DOMAIN of values to -1*Pi to +1*PI you see all there is to see. All the rest can be obtained by translation of the curve.

The RANGE of the sine function is LIMITED to values in the interval [-1, 1]

Let us summarize: The DOMAIN of the sine function is ]-infinity, +infinity[ and its RANGE is [-1,+1].

That being said, there is something I would like to point to you

These are the numbers.

You want a "square", so be it. Here is the window setting

and the corresponding picture. Does it look like a square?

Why do you insitst on drawing a square? Horizontally you have the angle ( a number with a unit), while vertically you have a ratio of two lengths ( a pure number). Would even think about a square if you drew your sine function with the degree as angle unit. Horizontally you would have a domain [-180 degrees, 180 degrees] while vertically you have a range [-3.14..., +3.14...]. How can that be a square?

I showed you how you can fix every dimension in the graph window (see the first picture) . Choose any values that you believe will give you a square graph. And I do mean to say "that make you believe", because there is no meaning attached to the "fact" that the window looks like a square. An angle cannot be compared to a the projection of one side of a right triangle onto the hypotenuse.

Feb 25, 2010 | Casio CFX-9850G Plus Calculator

I am enclosing a link to a web page that gives you the definition of the domain of a function and its range. You should read it to correct the confusion you entertain between domain and range.

In what follows I will speak of X and Y values.

Let me restate what I think you wanted to do. Ignore my rambling if you do not recognize your intention.

You want to draw the sin(X) function for X varying from Xmin= -2*pi radians to Xmax=2*pi radians. However, it you do not specify the Ymin and Ymax values for the vertical window dimensions a big portion of the screen is not used (vertically) and you want the height of the curve to take most of the screen. Specifically you want to set the vertical height of the window to go from Ymin=-1 to Ymax=1.

Press [SHIFT] [F3:V-Window] to set window dimensions.

On line Xmin= type : (-)2*[SHIFT][EXP] (Pi); or just (-) 6.3

Press [EXE] to validate the choice

On line Xmax= type 2*[SHIFT][EXP] (Pi) or just 6.3

Use down arrow to highlight the line Ymin= and enter (-)1

On line Ymax= type 1

You may or may not want to change the X(scale) and the Y(scale)

Press [EXIT] to exit the window settings.

Draw your graph.It will occupy all the screen.

Here is what you will get.

In what follows I will speak of X and Y values.

Let me restate what I think you wanted to do. Ignore my rambling if you do not recognize your intention.

You want to draw the sin(X) function for X varying from Xmin= -2*pi radians to Xmax=2*pi radians. However, it you do not specify the Ymin and Ymax values for the vertical window dimensions a big portion of the screen is not used (vertically) and you want the height of the curve to take most of the screen. Specifically you want to set the vertical height of the window to go from Ymin=-1 to Ymax=1.

Press [SHIFT] [F3:V-Window] to set window dimensions.

On line Xmin= type : (-)2*[SHIFT][EXP] (Pi); or just (-) 6.3

Press [EXE] to validate the choice

On line Xmax= type 2*[SHIFT][EXP] (Pi) or just 6.3

Use down arrow to highlight the line Ymin= and enter (-)1

On line Ymax= type 1

You may or may not want to change the X(scale) and the Y(scale)

Press [EXIT] to exit the window settings.

Draw your graph.It will occupy all the screen.

Here is what you will get.

Feb 24, 2010 | Casio CFX-9850G Plus Calculator

The inverse sine is only defined for values in the range [-1..+1]. Arguments outside this range will produce domain errors.

Apr 14, 2009 | Texas Instruments TI-83 Plus Calculator

Arcsine is only defined for arguments in the range [-1...+1]. If you supply an argument outside that range, you will get a domain error.

Mar 08, 2009 | Texas Instruments TI-84 Plus Calculator

Feb 22, 2018 | HP Office Equipment & Supplies

325 people viewed this question

Usually answered in minutes!

×