Using the online graphing calculator

lookagain said:
You want to graph \(\displaystyle g^{-1}(x) \ \ \), the inverse of g(x), at that site?

Can you type a specific g(x) function in this thread first for consideration to graphing

its inverse, \(\displaystyle g^{-1}(x), \) at that site?
Click to expand...

g(x) = 2(x + 3)^2 - 8 (- 3 < x < 0)

I also would like to know how I find the image set from the above using the domian above. I have looked into the coursebook today and can confidently say that no mathing examples can be found to this type of problem, although on this forum we have had a good discussion about tranlations and scaling this type of problem.


Now I know the above vertex would be ( - 3, - 8), and the roots would be x = - 1 and x = - 5, but without examples to work to I am struggling to understand how the roots - 1, and - 5 would be IF I am correct here in saying this, would be the domian of the above quadratic, then the domain (- 3< x < 0) is given as a shorter domain to work out the inverse?


I thought if I could have seen a graph first that I might of been in with half a chance of understanding but until I have some examples etc I am lost with it?


Thanks
 
Your "graphing calculator" graphs functions. It will not automatically find inverse functions for you- you have to do that yourself. You are given \(\displaystyle g(x)= 2(x+ 3)^2- 8\) for x between -3 and 0. You don't really need to know about "completing the square" because, here, the square is already "completed" for you. Write \(\displaystyle y= 2(x+ 3)^2- 8\) and "swap" x and y: \(\displaystyle x= 2(y+ 3)^2- 8\). That really is the "inverse" part- reversing x and y. But now you want to solve for y so you can write it in "function" form
To solve \(\displaystyle x= 2(y+ 3)^2- 8\) you have to "undo" what has been done to y. That is, if you were given a value for y, you would have to (1) add 3, (2) square, (3) multiply by 2, (4) subtract 8.

To solve we must do the opposite: the opposite of each step in the oppsite order. So

1) Add 8 to both sides: \(\displaystyle x+ 8= 2(y+ 3)^2\)
2) Divide both sides by 2: \(\displaystyle (1/2)(x+ 8)= (y+ 3)^2\)
3) Take the square root of both sides. This is where we need the "x between -3 and 0". A square has two possible inverses:
\(\displaystyle \sqrt{(1/2)(x+ 8)}= \pm\sqrt{(y+ 3)^2}= \pm (y+ 3)\)
Because the "x" in the original function has to be between -3 and 0, and we have "swapped" x and y, y must be between -3 and 0 here. A square root is always positive so since y> -3, y+ 3> 0. We must use the "+":
\(\displaystyle \sqrt{(1/2)(x+ 8)}= y+ 3\)
4) Substract 3 from both sides:
\(\displaystyle \sqrt{(1/2)(x+ 8)}- 3= y\)

So \(\displaystyle g^{-1}(x)= \sqrt{(1/2)(x+ 8)}- 3\)
Graph that.
 
HallsofIvy said:
Your "graphing calculator" graphs functions. It will not automatically find inverse functions for you- you have to do that yourself. You are given \(\displaystyle g(x)= 2(x+ 3)^2- 8\) for x between -3 and 0. You don't really need to know about "completing the square" because, here, the square is already "completed" for you. Write \(\displaystyle y= 2(x+ 3)^2- 8\) and "swap" x and y: \(\displaystyle x= 2(y+ 3)^2- 8\). That really is the "inverse" part- reversing x and y. But now you want to solve for y so you can write it in "function" form
To solve \(\displaystyle x= 2(y+ 3)^2- 8\) you have to "undo" what has been done to y. That is, if you were given a value for y, you would have to (1) add 3, (2) square, (3) multiply by 2, (4) subtract 8.

To solve we must do the opposite: the opposite of each step in the oppsite order. So

1) Add 8 to both sides: \(\displaystyle x+ 8= 2(y+ 3)^2\)
2) Divide both sides by 2: \(\displaystyle (1/2)(x+ 8)= (y+ 3)^2\)
3) Take the square root of both sides. This is where we need the "x between -3 and 0". A square has two possible inverses:
\(\displaystyle \sqrt{(1/2)(x+ 8)}= \pm\sqrt{(y+ 3)^2}= \pm (y+ 3)\)
Because the "x" in the original function has to be between -3 and 0, and we have "swapped" x and y, y must be between -3 and 0 here. A square root is always positive so since y> -3, y+ 3> 0. We must use the "+":
\(\displaystyle \sqrt{(1/2)(x+ 8)}= y+ 3\)
4) Substract 3 from both sides:
\(\displaystyle \sqrt{(1/2)(x+ 8)}- 3= y\)

So \(\displaystyle g^{-1}(x)= \sqrt{(1/2)(x+ 8)}- 3\)
Graph that.
Click to expand...

Halls

This is an old thread that was hijacked by a spam artist.
 
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