Can someone help me with power functions?

Question

javk · Answer

Let 'X' denote the set of non negative real numbers. Then the power function $$f:X \rightarrow X$$ defined by $$f(x)=x ^{s}$$ is a one-to-one correspondence for every positive rational number 's'. Hence each power function has an inverse function of 
$$f(x)=x$$            $$f(x)=x ^{\frac{ 1 }{ 4}}$$               $$f(x)=x ^{\frac{ 3 }{ 2 }}$$

anonymous · Answer

What exactly is your question? Are you looking for the inverse functions of each given power function?

javk · Answer

sorry...yeah my bad...I'm supposed to find the inverse functions of each

javk · Answer

so the first one would be:
$$f ^{-1}(x)=\sqrt[3]{x}$$
the second one would be:
$$f^{-1}=x^{4}$$
and the third one:
$$f^{-1} (x)= x^{\frac{2}{3}}$$
Correct?

javk · Answer

My biggest issue at the moment is with the words `power function`. When I see those words everything else just seemingly disappears into the background and these two words just stare back at me mockingly. Now if the question was phrased differently i.e. Find the inverse functions of the following functions, I should have had no issues whatsoever. 

Do you think you could perhaps help me out?

anonymous · Answer

You can think of power functions as expressions raised to a certain power. More generally, you might be given a polynomial, like $x^2+3x$, but these differ from power functions in the sense that polynomials are linear combinations of power functions.

Fortunately, power functions are fairly easy to work with when it comes to finding their inverses. In general, for the power function $f(x)=x^n$, the inverse is indeed $f^{-1}(x)=x^{1/n}$. This is because for any set of inverse functions, you have the relation 
$$f(f^{-1}(x))=x=f^{-1}(f(x))$$
Notice that
$$\left(x^n\right)^{1/n}=x^{n/n}=x\
\left(x^{1/n}\right)^n=x^{n/n}=x$$

anonymous · Answer

One thing to make note of is that not every power function has a perfect inverse. Consider the square function, or $f(x)=x^2$. This function is not one-to-one (take $x=1$ and $x=-1$, and you get the same value: $f(\pm1)=(\pm1)^2=1$), so it doesn't have an inverse.

However, you can restrict the domain such that it DOES have an inverse. In particular, we want to remove the values of $x_0$ that give use repeats of $f(x_0)$.

For $f(x)=x^2$, it would suffice for use to use the non-negative real numbers as our domain, i.e. $f:\mathbb{R}^+\cup\{0\}\to\mathbb{R}^+\cup\{0\}$, where $\mathbb{R}$ is the set of real numbers, $\mathbb{R}^+$ is the set of positive real numbers, and $\mathbb{R}^+\cup\{0\}$ is the set of positive reals plus 0.

For any power function $f(x)=x^n$ with even $n$, this sort of domain restriction will work nicely to get an appropriate $f^{-1}(x)$.

javk · Answer

So that means that the function of x is x to the power of a particular number is a power function?
So in the power function, the exponent is fixed but the base changes as per the value of x?

javk · Answer

One more thing would I have to limit the domain of every other even number exponent as well, for the same reasons?

anonymous · Answer

Yes, yes, and yes. Though nothing is stopping you from using all nonpositive real numbers. You just can't have both positive and negative reals.

javk · Answer

Thank you so much.

anonymous · Answer

yw