Squaring a real valued function

Question

There is a function f(x) defined as $$f(x)= \sqrt{x+1}$$ And we need to find its square that is $$f^2(x)$$ Or in other words we have to find $$f(x) × f(x)$$ What I am doing is :- First I found the domain of f(x) that is $$[-1,\infty)$$ Then I found the intersection of domains of f(x) and f(x) that is also equal to $$[-1,\infty)$$ And then we can write as :- $$f^2:[-1,\infty)\to$$ R is defined by $$f^2(x)=\sqrt{x+1}×\sqrt{x+1}$$ $$f^2(x)=x+1$$ My question is that in $f^2(x)$ we calculated the domain as $[-1,\infty)$ but we get $f^2(x)=x+1$ where we can put any real number so its domain should be R(real numbers). What am I misssing here??? Any help is highly appreciated.

Answer 1

Yes, the square of the function $f$ is the restriction of $x+1$ to $[-1,\infty)$. It can be analytically extended to all of $\mathbb R$, but the function itself, if defined as $f^2$, is defined on $[-1,\infty)$.

Answer 2

I think you are missing the fact that $x\mapsto f(x)^2= x+1$ on $[-1,\infty)$ is not $x\mapsto f(x)^2$ on $\mathbb R$. As long as your object $f$ or $f^2$ (which depends on the chosen domain and codomain!) satisfies the requirements to be a function you are fine. You could work with $f:(0,T]\to\mathbb R$, $x\mapsto \sqrt{x+1}$ for instance. No need to consider/extend functions to the biggest possible domain.

Answer 3

If $$f(x)=\sqrt{x+1}\space\text{for} \space x\in[-1,\infty)$$ Then $$f^2(x)=x+1\space \text{for}\space x\in[-1,\infty)$$ Obviously, we know $g(x)= x+1$ CAN take all $x \in \Bbb R$, however given that $f(x)$ is limited, so too is $f^2(x)$.

See this graph to see how it happens.

Answer 4

Since your first function $f(x)$ is defined over $[-1,\infty)$ it means that $x\in [-1,\infty)$, so when you define the new function $f^2(x)$, that function has the same domain as the original $x$ unless you redifine the domain of that function to $R$ for example. You'll see in calculus that you can redifine a rational function like $\frac{x^2-9}{x-3}$ like this \begin{cases} x+3, & \text{for } x \neq 3 \\ some\,value, & \text{for } x=3 \\ \end{cases}

so you don't need to define a rational function in that case for example.

Answer 5

A lot of this is convention. But let's look carefully at what's going on anyway. Let

$$g(x) = \sqrt{x+1} \times \sqrt{x+1}$$

Then $g(-2)$ is undefined according to the usual convention for domains of functions defined by formulas (when only real numbers are considered).

Now it just so happens that for each $x \geq -1$ we have

$$g(x) \; = \; \sqrt{x+1} \times \sqrt{x+1} \; = \; x+1$$

Note that this equation does not hold for any values of $x$ that are less than $-1,$ since the left hand side is not even defined for any such value of $x.$

Therefore, for each $x \geq -1$ we have

$$g(x)=x+1$$

and for each $x < -1$ we have

$$g(x) \;\; \text{is undefined.} $$

Now let's suppose someone who hasn't seen any of this asks us to consider the function

$$h(x) = x + 1, $$

and nothing is said about the domain of $h.$

Both that person and we would agree, according the the usual convention for domains of functions defined by formulas, that $h(x)$ is defined for all real numbers $x.$ However, it would be incorrect to say that $g$ and $h$ are the same function, even though they are defined by using the same formula, because $g$ has a special restriction ABOVE AND BEYOND THE USUAL CONVENTION FOR DOMAINS OF FUNCTIONS DEFINED BY FORMULAS, namely $g(x)$ is only defined for values of $x$ greater than or equal to $-1,$ whereas $h(x)$ is defined for all real numbers $x.$

For a similar but more extreme example, consider the functions

$$G(x) \; = \; 0 \cdot \sqrt{-x^2 - 1} $$ and

$$ H(x) = 0 $$