Well, remember we can't take the square root of a negative
so we see that we have

so find those values that take sqrt of a negative and restrict hem from the domain
anny value greater than 1 and less than -1
so domain is from -1 to 1, including those numbers
D=[-1,1]
a. D=[-1,1] or from -1 to 1 is domain
b. for a TI-84, go to y-editor then input

for y1
c. for a TI-84, click 2nd then window (gets to tbset) scrol down to set Δx to 0.1, then cilick 2nd again then click graph (to select table) and scroll down till you see that value of y that is the biggest, that value is x=0.7
A. domain is from -1 to 1
B. use your brain or google the instructions for your calulator
C. at x=0.7
Answer:
The inverse for log₂(x) + 2 is - log₂x + 2.
Step-by-step explanation:
Given that
f(x) = log₂(x) + 2
Now to find the inverse of any function we put we replace x by 1/x.
f(x) = log₂(x) + 2
f(1/x) =g(x)= log₂(1/x) + 2
As we know that
log₂(a/b) = log₂a - log₂b
g(x) = log₂1 - log₂x + 2
We know that log₂1 = 0
g(x) = 0 - log₂x + 2
g(x) = - log₂x + 2
So the inverse for log₂(x) + 2 is - log₂x + 2.
Answer:
C
Step-by-step explanation:
Using the sine ratio in the right triangle.
sinC =
=
=
, thus
∠ C =
(
) ≈ 34.85° ( to 2 dec. places )