This might be silly, but if $f(\sqrt{x})=\frac{0.1}{a}x$, is $f(x)=\frac{0.1}{a}x^2$?
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5If $f(\sqrt{x})=\frac{0.1}{a}x$ holds for all $x \geq 0$, then we have $f(x)=\frac{0.1}{a}x^2$ for all $x \geq 0$ but you cannot tell anything about $f(x)$ for $x<0$. – Sangchul Lee Aug 02 '20 at 19:16
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3Yes, with the restriction that $x\ge 0$. – Vishu Aug 02 '20 at 19:16
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Got it, thanks for the quick reply. – PGupta Aug 02 '20 at 19:18
1 Answers
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The argument of a function is a dummy variable, meaning that it is a placeholder for a number. Consider $g(x)=x^2$; we could just as equally have written $g(y)=y^2$. The function is about the operation performed on the argument, not the argument itself. It might help to think of $f$ as
$$ f(\text{number})=(\text{number})^2\times\frac{0.1}{a} $$
This makes it clearer what is going on: you take a number, you square it, and you multiply it by $\frac{0.1}{a}$. It is this kind of thinking that leads us directly to the answer:
$$ f(x)=x^2 \times\frac{0.1}{a}=\frac{x^2}{10a}$$
So remember, it is the rule that is the defining property of a function. The dummy variable is merely there to illustrate what happens to a number when it is inputted into the function. With this in mind, $\sqrt{x}$ seems like an odd choice. It is clearer, and less likely to cause confusion, to use something like $x$ or $y$ instead.
NB as Sangchul Lee has pointed out, the domain of the function is important: this answer assumes that $x\geq0$, as I presume you only care about real-valued functions.
Joe
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