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Mathematics 22 Online
OpenStudy (lncognlto):

I'm having a problem trying to find the inverse function of g(x) = 4(x - 3)^2 - 25.

OpenStudy (lncognlto):

I get \[g ^{-1}(x) = \frac{ 1 }{ 2 }\sqrt{x + 25} + 3\]

OpenStudy (lncognlto):

But apparently the answer is \[g ^{-1}(x) = 3 - \frac{ 1 }{ 2 }\sqrt{x + 25}\]

OpenStudy (lncognlto):

I don't understand why the answers are different?

OpenStudy (dobby1):

this wolf pick is even better than the last nice

OpenStudy (lncognlto):

Thanks, lol

OpenStudy (lncognlto):

@hartnn

OpenStudy (lncognlto):

@Luigi0210

hartnn (hartnn):

where is it mentioned that the answer is 3 -1/2 sqrt.... and NOT 3+1/2 sqrt ... ?

hartnn (hartnn):

actually , when you take the square root, you need to consider both positive and negative to get \(\pm \sqrt{x+25}\) so actually the final answer will be \(g ^{-1}(x) = 3 \pm \frac{ 1 }{ 2 }\sqrt{x + 25}\)

OpenStudy (lncognlto):

Yeah, that's what I thought. But it says in the answer scheme that 3 -1/2 sqrt.... is the answer. I am busy writing a past CIE exam paper as practice for finals.

hartnn (hartnn):

there's a domain mentioned with the answer, that was the criteria to rule out the other answer

OpenStudy (lncognlto):

Sorry, I'm having a not very bright moment. How does the domain of g^-1(x) is less than or equal to -9 rule out the one answer?

OpenStudy (ybarrap):

Is your question 10 iii in the Mark scheme doc (not sure what that means)? If so, it says x >= -9, not less than or equal.

OpenStudy (lncognlto):

Ja, the question I am working on is 10 (iii). Yes, that's right, I just looked again. My brain seems to have gone on the blink for some reason. >.<

OpenStudy (anonymous):

personally i think the easier way for you to look at it is the g(x) is y so your problem would be y=4(x - 3)^2 - 25. so now do 4 times x and 4 times -3 then you. now set up your equation and it would be y=4x-12^2-25 so now you just switch the y and x and solve for y. so your equation would be x=4y-12^2-25 now solve for y!! hope that helped

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