Help with graphing problem??

f(3)  = 3 - √[3 - 2]  =  3 - √1  = 3 - 1 = 2

Sorry,  ihavelovedyousincewewere18.....I just don't understand part e.

f(x) = x - √[x - 2]

So f(11) just means to put 11 into the function and evaluate the result......so we have

f(11) = 11 - √[11- 2]   =   11 - √9  =   11 - 3  = 8

f(6) , f(2)  and f(0) are done similarly

b.   Here's the graph.......https://www.desmos.com/calculator/lok5hw3jkm

c.  Note that x cannot be less than 2, because √[x - 2]  would result in taking the square root of a negative number and this isn't a "real" result.....so.....the domain is x ≥ 2

d.   x - √[x - 2 ]  = 0      add  √[x - 2 ]  to both sides

x =  √[x - 2 ]    square both sides

x^2 = x - 2      rearrange

x^2 - x + 2  = 0     Note that, in the quadratic formula.....b^2 - 4ac = (-1)^2 - 4(1)(2) =  1 - 8  = -7........and this would mean that we would have two non-real solutions......thus...... x - √[x - 2 ]  = 0   doesn't exist in real terms.......also, look at the graph.......the function never touches or crosses the x axis.....thus, it is never 0

f(6)  = 6 - √[6 - 2]  =  6 - √4  = 6 - 2 = 4

f(2) = 6 - √[2 - 2] = 6  - 0 = 6

f(0) = not defined

Woops, I totally made a typo. Its supposed to be 

a. find f(11), f(6), f(3), f(2), and f(0).

Does that change anything from f(3) on?

and what should f(3) be?

Also, I forgot to add one last question. 

e. Set f(x)=f and solve for x. Show on your graph that one solution is valid and the other one is extraneous. (Plot 2 points showing one is on the graph and the other is not.)