# The function $g$ is such that $g(x) = 3x^2 - 12$ for $x <q$, find the greatest possible value of $q$.

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Thanks to Mirko, now we have a complete statement of Q10:

The function $f$ is such that $f(x)=2x+3$ for $x\geq 0$. The function $g$ is such that $g(x)=ax^2+b$ for $x\leq q$, where $a$, $b$ and $q$ are constants. The function $f\circ g$ is such that $(f\circ g)(x)=6x^2−21$ for $x\leq q$. (i) Find the values of $a$ and $b$. (ii) Find the greatest possible value of $q$.  It is now given that $q=-3$. (iii) Find the range of $f\circ g$. (iv) Find an expression for $(f\circ g)^{-1}(x)$ and state the domain of $(f\circ g)^{-1}$.

OP solved Q10(i) and found that $g(x)=3x^2-12$.

Q10(ii) is asking the greatest possible value of $q$ such that $f(g(x))=6x^2-21$ is defined for $x\leq q$ (here $f(x)=2x+3$ is defined for $x\geq 0$). Therefore the composition is possible if $g(x)\geq 0$. Hence we should find the greatest possible value of $q$ such that $g(x)\geq 0$ FOR ALL $x\leq q$.

By solving the inequality we get that $g(x)\geq 0$ iff $x\in (-\infty,-2]\cup [2,+\infty)$. So what is $q$?

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### user450199

Updated on December 20, 2022

• user450199 11 months

This question is from CIE A Level Maths (9709) May/June 2016 Q10ii. The answer is $q = -2$, but I don't get how it is done. Anyone, please help me and explain to me. Thank you.

• Hagen von Eitzen over 6 years
Nothing prevents $g(x)=3x^2-12$ for all $x\in \Bbb R$, so $q$ is not bounded in any way
• DMcMor over 6 years
Surely there is more to the question than that.
• JB1 over 6 years
• fleablood over 6 years
The question makes no sense. You can set q to anything and say the function is that for any value of x less than q. There is no restriction or criterion. The question simply makes no sense.
• Toby Mak over 6 years
There are so many past papers to search for in that year and season. Why don't you specify what paper it is?
• Mirko over 6 years
could you find a link to your question? There are too many "May/June 2016 Q10ii" one of them is as 8mundo.com/file?id=3353147 and seems to come close to the statement of your question, but does not seem to be exactly your question. There is a lon glist at 8mundo.com/post/…
• Robert Z over 6 years
@Mirko I think that you found the right question paper!
• Mirko over 6 years
The correct statement is now a comment to the correct answer ... of course the OP could have provided it, and formatted it properly, hopefully this will happen in their future posts
• Mirko over 6 years
you mean OP found $g(x)=3x^2-12$, not $g(x)=3x^2+12$. I see now what they mean in that paper, indeed it is the right paper, where by $fg(x)$ they mean composition, $(f\circ g)(x)$, it makes sense now. (And of course $q=-2$ the OP already knew that :)
• Robert Z over 6 years
@ Mirko yes, thanks
• Mirko over 6 years
To make it self-contained, here is the statement. The function $f$ is such that $f(x)=2x+3$ for $x\ge0$. The function $g$ is such that $g(x)=ax^2+b$ for $x\le q$, where $a,b$ and $q$ are constants. The function $fg$ is such that $fg(x) = 6x^2−21$ for $x\le q$. (i) Find the values of $a$ and $b$. , (ii) Find the greatest possible value of $q$. , It is now given that $q=−3$. (iii) Find the range of $fg$. , (iv) Find an expression for $(fg)^{−1}(x)$ and state the domain of $(fg)^{−1}$, .