Question

How do you write an equation in the form $f\left( x \right)=k{{x}^{n}}$ for the direct variation functions given $f\left( 2 \right)=8$ and $n=4$?

Answer 1

Hint: The function $f\left( x \right)=k{{x}^{n}}$ for the direct variation functions can be expressed in the expression of $g\left( ax \right)=a\times g\left( x \right)$ for arbitrary function $y=g\left( x \right)$. Then we apply the indices theorem to get where ${{a}^{n}}$ can be expressed as the number $a$ with the exponent being $n$. Then we have the value of the k.

Complete step-by-step solution:
We first need to explain the meaning of a function being in direct variation. To understand it better we take a function where $y=g\left( x \right)$. Now we change the value of the input from $x$ to $ax$ where $a$ is a constant, then if we get $g\left( ax \right)=a\times g\left( x \right)$, we can say that the function $f\left( x \right)=k{{x}^{n}}$ is in variation.
For our given function $f\left( x \right)=k{{x}^{n}}$, the value of k is unknown. We need to find its value.
We have been given the entries where $f\left( 2 \right)=8$ and $n=4$.
Now, putting $x=2,n=4$ in $f\left( x \right)=k{{x}^{n}}$, we get $f\left( 2 \right)=k\times {{2}^{4}}$.
The equations $f\left( 2 \right)=8$ and $f\left( 2 \right)=k\times {{2}^{4}}$ gives $k\times {{2}^{4}}=8$.
Now we apply the indices formula.
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$.
The simplified form of the expression ${{a}^{n}}$ can be written as the multiplied form of number $a$ of n-times.
Therefore, ${{a}^{n}}=\underbrace{a\times a\times a\times ....\times a\times a}_{n-times}$.
So, ${{2}^{4}}=2\times 2\times 2\times 2=16$.
This gives
$\begin{align}
  & k\times {{2}^{4}}=16k=8 \\
 & \Rightarrow k=\dfrac{8}{16}=\dfrac{1}{2} \\
\end{align}$
Therefore, the equation is $f\left( x \right)=k{{x}^{n}}=\dfrac{{{x}^{n}}}{2}$.

Note: We need to remember that addition and subtraction for exponents works for taking common terms out depending on the values of the indices.
For numbers ${{a}^{m}}$ and ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$.the relation is independent of the values of $m$ and $n$.