# Ugent math as help!!!!

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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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#2

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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!

**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!

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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

I then did 1/x + 9 >= 0

and git x to be greater than or equal to -1/9

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#4

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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

**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

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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#7

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Also, the domain in the mark scheme is X > 0

**sweetescobar**)Also, the domain in the mark scheme is X > 0

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#8

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But when dividing by a negative, we reverse the sign correct?

**sweetescobar**)But when dividing by a negative, we reverse the sign correct?

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#10

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Im sorry, I don't follow

**sweetescobar**)Im sorry, I don't follow

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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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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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#13

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What does it mean when it says that function f and g are defined for x > 3? Is that the range?

**sweetescobar**)What does it mean when it says that function f and g are defined for x > 3? Is that the range?

https://www.mathsisfun.com/definitio...-function.html

Last edited by mqb2766; 1 month ago

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#14

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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)

**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)

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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(Original post by

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.

**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.

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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#16

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In the part where you say to 'assume x > 0 and solve. Then assume x < 0 and solve', why is that?

**sweetescobar**)In the part where you say to 'assume x > 0 and solve. Then assume x < 0 and solve', why is that?

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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#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.

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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Ok, but I don't understand why we are assuming x < 0 or greater than

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#20

(Original post by

Ok, but I don't understand why we are assuming x < 0 or greater than

**sweetescobar**)Ok, but I don't understand why we are assuming x < 0 or greater than

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