Question

If $f\left( x \right)={{x}^{2}}-x$, how do you find $3f\left( x \right)$?

Answer 1

Hint: We first express the function $f\left( x \right)={{x}^{2}}-x$ term wise. Then we explain that multiplication for such function works in term wise multiplication. We multiply 3 with both ${{x}^{2}}$ and $-x$. At the end we add those two terms to get the final answer for $3f\left( x \right)$.

Complete step by step answer:
The given function $f\left( x \right)={{x}^{2}}-x$ is a quadratic equation. We need to find the value of $3f\left( x \right)$.
The constant multiplication with the function is equivalent to multiplying the constant with individual terms.
Multiplying 3 both sides of $f\left( x \right)={{x}^{2}}-x$, we get
$\begin{align}
  & f\left( x \right)={{x}^{2}}-x \\
 & \Rightarrow 3f\left( x \right)=3\left( {{x}^{2}}-x \right) \\
\end{align}$
Therefore, 3 gets multiplied with both ${{x}^{2}}$ and $-x$.
Multiplying 3 with ${{x}^{2}}$. We get $3\times {{x}^{2}}=3{{x}^{2}}$. Multiplying 3 with $-x$. We get $3\times \left( -x \right)=-3x$.
Both terms get added at the end.
Therefore, $3f\left( x \right)=3\left( {{x}^{2}}-x \right)=3{{x}^{2}}-3x$.
We now verify the result with use of values for x.
Let’s take the value of $x=2$.
First, we find the value of $f\left( x \right)={{x}^{2}}-x$ and then multiply 3.
$f\left( 2 \right)={{\left( 2 \right)}^{2}}-2=4-2=2$. Multiplying with 3 will get $3f\left( x \right)=3\times 2=6$.
Now we place the value of $x=2$ in the equation $3{{x}^{2}}-3x$ and get
$3{{\left( 2 \right)}^{2}}-3\times 2=3\times 4-6=12-6=6$.
Both the values are the same.

Thus, verified if $f\left( x \right)={{x}^{2}}-x$, then $3f\left( x \right)=3{{x}^{2}}-3x$.

Note: Not just multiplication but addition, subtraction and division for constants all works term wise. But if we change from constants to variables then only addition and subtraction can be completed term wise. Division and multiplication become complicated to operate.