Let $f : [0,3] \rightarrow B$ be defined by $f(x) = 4x-(x^2+1)$. Determine the image domain (or maybe it's called range in English?) B so that the function becomes surjective. Does it become invertible?
How should I approach this? I've been thinking:
By graphing the function, I can see that the lowest value is when $f(0) = -1$ and the highest is when $f(2) = 3$.
To make the function surjective, I need to map every value from the domain to the range. Therefore, $B = [-1,3]$.
Since the function is a parabola, it is not invertible.
So, $f : [0,3] \rightarrow [-1,3], f(x) = 4x-(x^2+1)$ is surjective but not invertible.
Am I wrong? I have nothing to compare my answer to so I don't really know...