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Power functions
Let #a=\frac{p}{q}# be a rational number that is not an integer, with integers #p# and #q# unequal to zero and #q\gt0#. From the chapter Numbers it became clear that #x^a# is a function of #x# with domain #\ivco{0}{\infty}# if #a\gt0# and #\ivoo{0}{\infty}# if #a\lt0#. After all, #x^a = \sqrt[q]{x^p}# is the unique number #b\ge0# for which #b^q =x^p# applies.
Even if #a# is not rational, we can define #x^a# on #\ivco{0}{\infty}# if #a\gt0# and on #\ivoo{0}{\infty}# if #a\lt0#. Through numbers of the form # \sqrt[q]{x^p}#, with #\frac{p}{q}# a rational number close to #a#, the number can be approximated as well as needed. Technically, this is a bit more difficult than the rational case.
Power function
Let #a# be real number. The function #x^a# of #x# is called the power function with exponent #a#.
If #a=0#, then #x^a# is the constant function, except that it is not defined in #0#.
The domain can be chosen as in the case where #a# is rational.
If #a\gt0#, then it would make sense to state #0^a =0#, consequently increasing the domain of the power function to #\ivco{0}{\infty}#. However, in order not to jeopardize the rules of calculation for powers below, we keep the domain #\ivoo{0}{\infty}#.
The function #x^a# of #x# with domain #\ivoo{0}{\infty}# has range #\ivoo{0}{\infty}#.
If #a\gt0# and if we take the domain of #x^a# to be #\ivco{0}{\infty}#, then the range would become #\ivco{0}{\infty}#.
All known rules for fractional powers also apply here.
Rules of calculation for powers
Let #a# and #b# be real numbers and #x# and #y# positive numbers. Then the following equalites apply:
- #\left(x^a\right)^b = x^{a\cdot b}#
- #\left(x\cdot y\right)^a = x^a\cdot y ^a#
- #x^a\cdot x^b = x^{a+b}#
- #\frac{x^a}{x^b}=x^{a-b}#
- #x^0 = 1#
These laws are familiar from the theory Fractional powers in the case where #a# and #b# are rational numbers.
These rules lead to the following convenient properties of power functions. Here we will consider #x^a# as function with domain #\ivoo{0}{\infty}#. If #a# is a number not equal to zero, then the function is defined on #\mathbb{R}\setminus\{0\}# (all numbers except #0#), but then laws like #x^3 = x^{\frac{6}{2}} =\sqrt[2]{x^{6}}# no longer apply.
Let #a# be a real number, unequal to #0#. The power function #x^a# with domain #\ivoo{0}{\infty}#
- has range #\ivoo{0}{\infty}#,
- is increasing if #a\gt 0# and decreasing if #a\lt0#,
- is bijective
- has #x^{\frac{1}{a}}# as inverse.
Exclusion of #x=0# is needed, because for example #x^{-1}# is now present and equal to #\frac{1}{x}#.
Exclusion of #a=0# is needed, because #x^0# is the constant function #1#, and hence, not bijective.
About the proof:
The monotony (being increasing or decreasing) of #x^a# can be concluded from the same property in the special case where #a# is a rational number.
From Injectivity for monotonic functions it follows that #x^a# is injective on #\ivoo{0}{\infty}#.
The rules of calculation show that if #y=x^a#, then #x=y^{\frac{1}{a}}#, and the other way around. Consequently #x^{\frac{1}{a}}# is the inverse function of #x^a#.
We will be working towards an equation without roots.
\[
\begin{array}{rclcl}
-3x^{{{3}\over{2}}}&=&\dfrac{-3}{\sqrt{x}}&\phantom{xxx}&\color{blue}{\text{the given equation}}\\
-3x^{2}&=&-3&\phantom{xxx}&\color{blue}{\text{multiplying by }\sqrt{x}}\\
x^{2}&=&1&\phantom{xxx}&\color{blue}{\text{dividing by }-3\tiny.}\\
\end{array}
\]
The theory of Higher power roots states that there are two solutions, being: #x=1\lor x=-1#. Hence, these two values are solutions to the equation #x^{2}=1#, but not necessarily to the original equation #-3x^{{{3}\over{2}}}=\dfrac{-3}{\sqrt{x}}#. Since the original equation is only defined when #x\gt0#. Hence, the negative solution #x=-1# is not valid, and the only solution is #x=1#.
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