When finding the value of "x" what do I use in place of it?

example: graphing -5< 4x+3< 3

How do I put the "x" into the calculator?

Re: How do I get an 'x'?

Hello,

Use the key [x, theta, T] below the [Alpha] key, or use [ALPHA] [+] (x)

Hope it helps.

Posted on Sep 22, 2009

There are several ways of doing this.

Since you're using a graphing calculator, you can always graph it as a function of 'a' and use the graphical tools.

Another way is to use the solve() function. For example, to find the value of a where the integral of sin(x) from 0 to a equals one, enter

solve(fnInt(sin(X),X,0,A)-1),A,1.5)

solve( and fnInt( are accessible through the CATALOG. The 1.5 is an initial guess.

Since you're using a graphing calculator, you can always graph it as a function of 'a' and use the graphical tools.

Another way is to use the solve() function. For example, to find the value of a where the integral of sin(x) from 0 to a equals one, enter

solve(fnInt(sin(X),X,0,A)-1),A,1.5)

solve( and fnInt( are accessible through the CATALOG. The 1.5 is an initial guess.

Apr 19, 2014 | Texas Instruments TI-84 Plus Silver...

At first in Y editor type **Y1=2*cosx** and for example **y2=[5*exp(-x)]-5** and then graphing it. Set **RAD** mode of course. See captured images

Jul 10, 2012 | Texas Instruments TI-84 Plus Silver...

Make sure you have the angular mode set to radians. In degree, the graph of sine(x) is nearly a straight line for small values of x.

May 08, 2012 | Texas Instruments TI-Nspire CAS Graphing...

There is only one stationary point for the function **y(x)=((x-1)^(1/3))*((x+2)^(2/3))** and it is pointwith coordinates** x=0 and y=-1.5859**

At first, first derivation is**y'(x)=x/((x-1)^(2/3))*((x+2)^(1/3))** and we obtained that **y'(0)=0**. Then we can find second derivation of the function and for value x=0 we could concluded that **y''(0)>0**. In this stationary point it is minimum value of the function.

Finally, at the points x=-2 and x=1 we have so called**critical values** for this function.

See captured images below

1. Second derivation for y(x)

2. Graph y(x)

3. Enlarged detail of the graph y(x)

At first, first derivation is

Finally, at the points x=-2 and x=1 we have so called

See captured images below

1. Second derivation for y(x)

2. Graph y(x)

3. Enlarged detail of the graph y(x)

Mar 27, 2012 | Texas Instruments TI-89 Calculator

You might not see the scatter plot when you go to the graph because the boundaries are not set correctly. With the scatter plot turned on, go to "zoom" and then scroll down to "zoomStat" and hit enter. Try going to the graph again and see if the the scatter plot shows up. As for the the equation being slightly different than the examples, this is probably just the calculator rounding up the answers it gets.
Good luck!

Jun 24, 2011 | Texas Instruments TI-84 Plus Silver...

If you want to graph the absolute value of a function, feed the function to the absolute value ie graph y= abs(f(x)). The calculator will take care of it. See screen captures where I draw the function y1=x^3-5x^2-3 and y2='x^3-5x^2-3'. On the first graph the function is represented and on the second the function and its absolute value.

The calculator knows how to calculate the absolute value of an expression. The name of the command is abs(

The calculator knows how to calculate the absolute value of an expression. The name of the command is abs(

May 08, 2011 | Texas Instruments TI-89 Calculator

simply divide the function by the domain you wish.

In your example, you wanted to graph y=x for x>=0.

You should graph the function f1=(x)/(x>0)

To find out when y=900, graph the function in f1, and then graph the equation y=900 (which will be a horizontal line) and use the intersect function to find where they cross. The x-coordinate of the intersection is the solution you are seeking.

In your example, you wanted to graph y=x for x>=0.

You should graph the function f1=(x)/(x>0)

To find out when y=900, graph the function in f1, and then graph the equation y=900 (which will be a horizontal line) and use the intersect function to find where they cross. The x-coordinate of the intersection is the solution you are seeking.

Jan 25, 2011 | Texas Instruments TI-Nspire Graphic...

Hello,

Xmin must be less than Xmax; Ymin must be less than Ymax. If your X variable takes only positive values you can set Xmin=0; if you want to graph y=X^2, you know that all y-values are going to be positive, so there is no need to waste screen real estate by putting Ymin negative.

It is like taking a picture with a camera. You point the viwefinder where the action is.

My suggestion is to set the default values Xmin=-10, Xmax=10, Ymin=-10, Ymax=10. Then look at the graph and locate the part that interests you more and focus on it by zooming in. Your knowledge of the general behavior of the function will help you determine what is worth looking at.

Hope it helps.

Xmin must be less than Xmax; Ymin must be less than Ymax. If your X variable takes only positive values you can set Xmin=0; if you want to graph y=X^2, you know that all y-values are going to be positive, so there is no need to waste screen real estate by putting Ymin negative.

It is like taking a picture with a camera. You point the viwefinder where the action is.

My suggestion is to set the default values Xmin=-10, Xmax=10, Ymin=-10, Ymax=10. Then look at the graph and locate the part that interests you more and focus on it by zooming in. Your knowledge of the general behavior of the function will help you determine what is worth looking at.

Hope it helps.

Sep 30, 2009 | Texas Instruments TI-84 Plus Calculator

Hello,

Here how you do it taking as exemples y1=x^3 -5 and y2=abs(x^3-5)

Turn calculator ON by pressing [AC/On] . Use arrows to select [Graph] icon and press [EXE]

On line Y1= enter X[^]3 -5 [EXE]

On line Y2= press [OPTN] [F5:NUM][F1:abs] [(] X [^]3 -5 [)] [EXE]

The graphs will be drawn one after the other. You might have to adjust the window limits.

Here is an ouput drawn on another calculator.

For negative values of the x-variable, Y1 has negative values, while in the same range, Y2 has positive values, equal and opposite to those of Y1. In that range, Y1 and Y2 are symetric of each other with respect to the X axis.

For positive values of the x-variable, Y1 and Y2 are identical.

Hope it helps

Here how you do it taking as exemples y1=x^3 -5 and y2=abs(x^3-5)

Turn calculator ON by pressing [AC/On] . Use arrows to select [Graph] icon and press [EXE]

On line Y1= enter X[^]3 -5 [EXE]

On line Y2= press [OPTN] [F5:NUM][F1:abs] [(] X [^]3 -5 [)] [EXE]

The graphs will be drawn one after the other. You might have to adjust the window limits.

Here is an ouput drawn on another calculator.

For negative values of the x-variable, Y1 has negative values, while in the same range, Y2 has positive values, equal and opposite to those of Y1. In that range, Y1 and Y2 are symetric of each other with respect to the X axis.

For positive values of the x-variable, Y1 and Y2 are identical.

Hope it helps

Jan 30, 2009 | Casio CFX-9850G Plus Calculator

If you have more than one function graphed, use Up/Down to select the one you want. Press Left/Right to select the point at which you want the derivative, then press ENTER to see the value of dy/dx.

Oct 06, 2008 | Texas Instruments TI-83 Plus Calculator

Oct 11, 2011 | Casio CFX 9850GA Plus Calculator

67 people viewed this question

Usually answered in minutes!

×