Question

How do you write an equation in the form $f\left( x \right) = k{x^n}$ for the direct variation function given $f\left( {0.3} \right) = 45$ and $n = 2$ ?

Answer 1

Hint: In order to solve this question, we need to find the constant of proportionality which is $k$. We do this by equating the function $f\left( x \right)$ with $k{x^n}$ . We place all the given values and then solve and simplify the sum to get our required answer.

Complete step-by-step solution:
In the given question, the given function is directly proportional which means
$f\left( x \right) \propto k{x^n}$, Where $k$ is the constant of proportionality.
Now according to the given, $f\left( {0.3} \right) = 45$and also $f\left( x \right) = k{x^n}$ and $n = 2$
Therefore, $f\left( {0.3} \right) = k{x^n} = 45$
Thus, $k{\left( {0.3} \right)^2} = 45$ ……. Through this, we can find the constant of proportionality.
$ \Rightarrow k\left( {0.09} \right) = 45$
Now we express the decimal part in fraction as:
$ \Rightarrow k \times \dfrac{9}{{100}} = 45$
Multiplying both sides with $\dfrac{{100}}{9}$, we get:
$ \Rightarrow k = 45 \times \dfrac{{100}}{9} = 500$

Therefore, $f\left( x \right) = 500{x^2}$

Note: In this question, the function was directly variable to a certain condition. This simply meant that if we were to change the value of this condition then the value of the function will also change similarly. It is directly variable or proportional meaning that if we were to increase the value of this condition then the numerical value of the function will also increase, and if we were to decrease the numerical value of the condition, then the numerical value of the function will also decrease.
On the other hand, inversely variable or inversely proportional means that if we were to increase the value of that particular condition, then the numerical value of our function will decrease and vice versa.
This proportionally is denoted by the constant of proportionality $k$
Directly proportional are represented as $ \propto k$
While inversely proportional are presented as $ \propto \dfrac{1}{k}$