Question

Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.

Answer 1

Injectivity:
Let x and y be two elements in the domain (R), such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \cos \left(x+2\right)=\cos \left(y+2\right) \\ \Rightarrow x+2=y+2\text{or}x+2=2\mathrm{\pi }-\left(y+2\right) \\ \Rightarrow \mathit{\text{x}}\text{=}\mathit{\text{y}}\text{or}\mathit{\text{x}}\text{+2=2π-}\mathit{\text{y}}\text{-2} \\ \Rightarrow \mathit{\text{x}}\text{=}\mathit{\text{y}}\text{or}\mathit{\text{x}}\text{=2π-}\mathit{\text{y}}\text{-4} \\ \mathrm{So},\mathrm{we}\mathrm{cannot}\mathrm{say}\text{that}\mathrm{x}=\mathrm{y} \\ \mathrm{For}\mathrm{example}, \\ \cos \frac{\mathrm{\pi }}{2}=\cos \frac{3\mathrm{\pi }}{2}=0 \\ \text{So,}\frac{\mathrm{\pi }}{2}\text{and}\frac{3\mathrm{\pi }}{2}\text{have the same image 0.} \end{gathered}$$
$\Rightarrow$ f is not one-one.
$\Rightarrow$ f is not a bijection.
Thus, f is not invertible.