Prove whether {f}:mathbb{R}tomathbb{R} text{defined by} f{{left({x}right)}}={4}{x}-{2} is a linear transformation. Prove whether {f}:mathbb{R}tomathbb

Step 1
To prove that $f(x)$ is linear transformation we have to prove that
$f(x+y)=f\left(x\right)+f\left(y\right)$
And af $\left(x\right)=f\left(ax\right),$ a belongs to R
Here a is a scalar and R is the set of real numbers.
Step 2
$f\left(x\right)=4x-2\text{and let}f\left(y\right)=4y-2$
Now we check if $f(x+y)=f\left(x\right)+f\left(y\right)$
$f(x+y)=4(x+y)-2=4x+4y-2$
And,
$f\left(x\right)+f\left(y\right)=4x-2+4y-2=4x+4y-4$
$\Rightarrow f(x+y)\text{is not equal to}f\left(x\right)+f\left(y\right)$
Hence $f\left(x\right)=4x-2$ is not a linear transformation.
Step 3
${\displaystyle f\left(x\right)=2x\text{and}{\displaystyle f\left(y\right)=2y}}$
${\displaystyle f(x+y)=2(x+y)=2x+2y}$
And,
${\displaystyle f\left(x\right)+f\left(y\right)=2x+2y}$
We get, ${\displaystyle f(x+y)=f\left(x\right)+f\left(y\right)}$
And,
af ${\displaystyle \left(x\right)=a\left(2x\right)=2ax=f\left(ax\right)}$
Both the properties are satisied, hence ${\displaystyle f\left(x\right)=2x}$ is a linear transformation.
Step 4
${\displaystyle f\left(x\right)=2xatx=1,f\left(1\right)=2}$
Hence ${\displaystyle f:R\to R,f\left(x\right)=2x\text{is equivalent to}{\displaystyle T:R\to R,T\left(1\right)=2.}}$