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Now, $f(x)$ is defined for all values of $x$ for which $x+2 \geq 0$
$x+2 \geq 0 \implies x \geq -2$

So, $\mathrm{Domain}(f)=[-2,\infty)$ which means $f : [-2,\infty) \longrightarrow \Bbb R$

$f^2(x) = \Big (f(x) \Big )^2=(\sqrt{x+2})^2=x+2$

So, $f^2$ is defined for all values of $x$, right? So, shouldn't $f^2:\Bbb R \longrightarrow \Bbb R$ ?

According to my textbook, $f^2:[-2,\infty) \longrightarrow \Bbb R$

But if we take something outside of $[-2,\infty)$, for example $-5$ and put it in $f^2(x)$, we get:
$f^2(-5) = \Big (f(-5) \Big )^2=(\sqrt{-5+2})^2=(\sqrt {-3})^2=(-3) \in \Bbb R$

Doesn't this mean that $f^2$ is defined for values outside the domain of $f$ as well? So, am I right or is the book right? If the book's right, where am I wrong?

Thanks!

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  • $\begingroup$ Your book is right. You can define a new function $h(x): \mathbb{R} \to \mathbb{R}$ s.t. $h(x):= x+2$ but $f^{2}(x)=(f(x))^{2} \neq h(x)$ If plug in $-5$ in $h$, then $h(-5) = -3$, but I simply can't plug in $-5$ in $f^{2}$ just because of the way it is defined. $\endgroup$ – Derpp May 15 '20 at 15:08
  • $\begingroup$ @Derpp I would say, slightly differently, that $f^2(x) = h(x)$ for all $x \ge -2$, but $f^2 \neq h$. $\endgroup$ – user786879 May 15 '20 at 15:13
  • $\begingroup$ Yes of course, $f^{2}$ and $h$ do agree on their values $\forall x \geq -2$ $\endgroup$ – Derpp May 15 '20 at 15:17
  • $\begingroup$ As a rule of thumb, the book is right. $\endgroup$ – Yves Daoust May 15 '20 at 15:19
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The textbook is correct. The domain of $f(x)=\sqrt{x+2}$ is $[-2,\infty)$ which therefore implies that the domain of $f^2(x)$ is also $[-2,\infty)$.

The new function that you constructed, $h(x)=x+2$, is the not the same as $f^2(x)$ since the domain of $h(x)$ is $(-\infty,\infty)$ while the domain of $f^2(x)$ is $[-2,\infty)$.

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  • $\begingroup$ But isn't the domain of a function the set of values for which it is defined. And since $f^2(x)$ is defined for all real numbers, why choose it's domain as the domain of $f$? $\endgroup$ – Rajdeep Sindhu May 15 '20 at 15:25
  • $\begingroup$ $f^2(x)$ isn't defined for all real numbers as $f^2(x)$ is restricted by the domain of $f(x)$. $\endgroup$ – Axion004 May 15 '20 at 15:26
  • $\begingroup$ So, if we have $f(x) = P$, where $P$ is some expression in terms of $x$, and $f^n(x) = Q$, where $Q$ is another expression in terms of $x$, and $h(x) = Q$, so $f \neq h$, since $Domain(f^n)$ is the same as the domain of $f$ since $f^n$ is a direct result of the definition of $f$ and $Domain(f) \neq Domain (h)$ which means that $Domain(f^n) \neq Domain(h)$ but the value of $h(y)=f^n(y) \forall y \in Domain(f^n) \cap Domain(h)$ and $Range(f^n) = range(h)$. Am I right? $\endgroup$ – Rajdeep Sindhu May 15 '20 at 15:35
  • $\begingroup$ Yes, although it would be $h(x)=f^n(x)$ and the two are the same when $n=1$. The domain of $f^n(x)$ would always be restricted by the domain of $f(x)$ (in terms of the square root function, it would always be when the quantity inside the square root is $\ge 0$ assuming that we are working in the real numbers.) $\endgroup$ – Axion004 May 15 '20 at 15:45
  • $\begingroup$ That clears it. Thanks again! $\endgroup$ – Rajdeep Sindhu May 15 '20 at 15:46
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You have written $f^2(-5)=(f(-5))^2$. It is correct, however, look at the inside function of the RHS. It is $f(-5)$. Can you define $f(-5)$? No. That means $f^2(-5)$ is undefined. Similarly $f^2$ is undefined for any $x<-2$. So, your book is correct.

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$f^2$ is the square of $f$. If $f$ is not defined, neither is $f^2$.

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  • $\begingroup$ This holds true just because $f^2$ is a result of $f$, right? Also, please check out my comment on @Axion004 's answer too... $\endgroup$ – Rajdeep Sindhu May 15 '20 at 15:39
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for example $-5$ and put it in $f^2(x)$, we get: $$f^2(-5) = \Big (f(-5) \Big )^2=(\sqrt{-5+2})^2 =(\sqrt {-3})^2=(-3) \in \Bbb R$$

(conclusion in) your answer is not that correct.

Assume you are given a function $f(x)=\frac{x}{x}$. What do you think is the domain of this? $\mathbb{R}$ (because it reduces to $f(x)=1$)? No! Actually the domain is $\mathbb{R} \setminus \{0\}$. And similarly the domain of $f(x)=(\sqrt{x})^2$ is $[0,+\infty)$ not $\mathbb{R}$

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$f^n(x)$ exists only in the domain of $f(x)$. This is because the value of $f(x) \notin R \space \forall \space x \notin D$, where $D$ is the domain. It is only a coincidence that squaring the complex number brings it to the real plane.

Here's a graph for verification. enter image description here

Notice how the domain of $g(x)$ is only $[-2,\infty)$ and not $R$. Thus, your book is correct in this case.

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  • $\begingroup$ I suppose you mean that $\forall x\not\in D$? $\endgroup$ – user12986714 May 15 '20 at 15:14
  • $\begingroup$ Yes indeed. Edited now. $\endgroup$ – Aniruddha Deb May 15 '20 at 15:15
  • $\begingroup$ So, we have defined the domain of the square of a function as the domain of that function and not as the values for which the function is defined, right? $\endgroup$ – Rajdeep Sindhu May 15 '20 at 15:26
  • $\begingroup$ @Rajdeep_Sindhu yes $\endgroup$ – Aniruddha Deb May 15 '20 at 15:32
  • $\begingroup$ @AniruddhaDeb Thanks a lot! $\endgroup$ – Rajdeep Sindhu May 15 '20 at 15:33

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