# Ugent math as help!!!!

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#1
Please help with me the second part of this question (Q9), I found the inverse function but don't and never know how to find the domain!
Last edited by sweetescobar; 1 month ago
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1 month ago
#2
(Original post by sweetescobar)
Please help with me the second part of this question (Q9), I found the inverse function but don't and never know how to find the domain!
what did you get for the inverse? As the question says "state", the domain should be reasonably straightforward.
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#3
I got the square root of 1/x + 9

I then did 1/x + 9 >= 0

and git x to be greater than or equal to -1/9
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1 month ago
#4
(Original post by sweetescobar)
I got the square root of 1/x + 9

I then did 1/x + 9 >= 0

and git x to be greater than or equal to -1/9
You may have a problem when x=0 with your expression? You really need to be a bit more careful with flipping an inequality.
Just think about the range of f() for x>3. Thats the domain of of the inverse.
Last edited by mqb2766; 1 month ago
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#5
But when dividing by a negative, we reverse the sign correct?
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#6
Also, the domain in the mark scheme is X > 0
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1 month ago
#7
(Original post by sweetescobar)
Also, the domain in the mark scheme is X > 0
I agree with that.
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1 month ago
#8
(Original post by sweetescobar)
But when dividing by a negative, we reverse the sign correct?
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#9
Im sorry, I don't follow
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1 month ago
#10
(Original post by sweetescobar)
Im sorry, I don't follow
Which part? As they say state, I guess they want you to just note the range of f() for x>3. That's fairly obviously y>0. Do you agree? Hence thats the domain of the inverse, so x>0.
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#11
What does it mean when it says that function f and g are defined for x > 3? Is that the range?
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#12
Also, I have tried to the values that are smaller than -1/9 in the inverse, and they would not work, only those larger would, with exception for x = 0. So, should the domain be: [-1/9,0) U (0,infinity)
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1 month ago
#13
(Original post by sweetescobar)
What does it mean when it says that function f and g are defined for x > 3? Is that the range?
It would be defined in your textbook, but that is the domain of f().
https://www.mathsisfun.com/definitio...-function.html
Last edited by mqb2766; 1 month ago
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1 month ago
#14
(Original post by sweetescobar)
Also, I have tried to the values that are smaller than -1/9 in the inverse, and they would not work, only those larger would, with exception for x = 0. So, should the domain be: [-1/9,0) U (0,infinity)
Did you try a value in (-1/9,0) or (-inf,-1/9) or both - what do you mean by smaller? Tbh, this is error spotting though. It would be easier to sketch the function and its inverse or plot in desmos.
https://www.desmos.com/calculator/qqoz2fcmq7

If you did the algebra to solve the inequality properly,
1/x + 9 >= 0
Assume x>0 and solve. Then assume x<0 and solve. What do you get in each case - upload your working pls? Which corresponds to the inverse function (domain)?

However, the question says state, so you need to be able to use the original function domain to work out the range and hence the domain of the inverse without any/much working for this question.
Last edited by mqb2766; 1 month ago
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#15
(Original post by mqb2766)
Did you try a value in (-1/9,0) or (-inf,-1/9) or both - what do you mean by smaller? Tbh, this is error spotting though. It would be easier to sketch the function and its inverse or plot in desmos.
https://www.desmos.com/calculator/qqoz2fcmq7

If you did the algebra to solve the inequality porperly,
1/x + 9 >= 0
Assume x>0 and solve. Then assume x<0 and solve. What do you get in each case - upload your working pls? Which corresponds to the inverse function (domain)?

However, the question says state, so you need to be able to use the original function domain to work out the range and hence the domain of the inverse without any/much working for this question.
In the part where you say to 'assume x > 0 and solve. Then assume x < 0 and solve', why is that?

Oh and yes, I think I figured it out from the domain of function f, being x > 3, since thats the range of f^-1, also, because the range of function f is x > 0, then that will be the domain for the inverse.

Thanks a lot for actually responding, I really appreciate it!
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1 month ago
#16
(Original post by sweetescobar)
In the part where you say to 'assume x > 0 and solve. Then assume x < 0 and solve', why is that?
You're solving
1/x + 9 >= 0
Write down the steps to do it clearly (and upload) and when you multiply or divide by x, you have to know the sign of x to decide whether to flip the inequality or not.
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#17
Here is my working
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1 month ago
#18
But what if
x < 0
Then the inequality should reverse. Indeed your "solution" has both positive and negative x, yet you didn't treat the inequality differently. So do what was suggested in the previous post about treating x>0 and x<0 seperately.

Try and relate it to the desmos picture. Your (-1/9,0) interval is impossible.
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#19
Ok, but I don't understand why we are assuming x < 0 or greater than
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1 month ago
#20
(Original post by sweetescobar)
Ok, but I don't understand why we are assuming x < 0 or greater than
Because you multiply through the inequality by x. You have to know its sign to decide whether to reverse the inequality or not. I've mentioned it several times.
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