Floor function

[TinaW]

Hi, I am in a bit of situation, professor is refusing to give us results and im quite sure that i got less points than should have. Please could somebody check few of the questions (ill post them separately as per rules) and see if i did correct solution?

 

[Dr.Peterson]

Hi, I am in a bit of situation, professor is refusing to give us results and im quite sure that i got less points than should have. Please could somebody check few of the questions (ill post them separately as per rules) and see if i did correct solution?

View attachment 19846
View attachment 19847

You seem to have totally ignored the floor function! That changes the range significantly.

And you seem to have answered the question about injective, etc., for the function f rather than g. That question is (b), not part of (a).

I can't quite read your image; please make the next one a little larger.

 

[pka]

\(\left\lfloor {\dfrac{{{{( - 1)}^2} + 1}}{3}} \right\rfloor = \left\lfloor {\dfrac{2}{3}} \right\rfloor \quad ,\left\lfloor {\dfrac{{{{(0)}^2} + 1}}{3}} \right\rfloor = \left\lfloor {\dfrac{1}{3}} \right\rfloor \quad ,\left\lfloor {\dfrac{{{{(1)}^2} + 1}}{3}} \right\rfloor = \left\lfloor {\dfrac{2}{3}} \right\rfloor \quad ,\left\lfloor {\dfrac{{{{(2)}^2} + 1}}{3}} \right\rfloor = \left\lfloor {\dfrac{5}{3}} \right\rfloor \)
What are the correct answers?

 

[TinaW]

Oh my! So mistake is in not using floor function (i followed professors solution with different values, he as well ignored floor function).

Edited because i couldnt see number, but was correct after all (sorry about small photo)