+130555 Here is ihavelovedyousincewewere18's original question for part e.....
e. Set f(x)=4 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.)
4 = x - √[x- 2] rearrange
√[x- 2] = x - 4 square both sides
x - 2 = x^2 - 8x + 16 subtract x and add 2 to both sides
x^2 - 9x + 18 = 0 factor
(x - 6) (x - 3) = 0 and setting both factors to 0, we have that x = 6 or x = 3
Notice that x = 6 is a valid answer to the original problem because 6 - √[6- 2] = 6 - √4 = 6 - 2 = 4
But x = 3 isn't valid because 6 - √[3- 2] = 6 - √1 = 6 - 1 = 5
Here's the graph........https://www.desmos.com/calculator/uakphwshii
Notice that (6, 4) is on the graph, but (3, 5) isn't
+130555 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
