Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.
Question:
Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.
Solution:
Injectivity:
Let x and y be two elements in the domain (R), such that
$f(x)=f(y)$
$\Rightarrow \cos (x+2)=\cos (y+2)$
$\Rightarrow x+2=y+2$ or $x+2=2 \pi-(y+2)$
$\Rightarrow x=y$ or $x+2=2 \pi-y-2$
$\Rightarrow x=y$ or $x=2 \pi-y-4$
So, we cannot say that $x=y$
For example,
$\cos \frac{\pi}{2}=\cos \frac{3 \pi}{2}=0$
So, $\frac{\pi}{2}$ and $\frac{3 \pi}{2}$ have the same image 0 .
$\Rightarrow f$ is not one-one.
$\Rightarrow f$ is not a bijection.
Thus, $f$ is not invertible.