Given f(x) = 2x^2- 1 with a domain d = {-1,0,1} find the range of f

Hello!

To find the range of f(x), we would need to substitute the given domain values into the given function and solve. Remember to use PEMDAS.

f(x) = 2x² - 1, x = -1

f(x) = 2(-1)² - 1

f(x) = 2(1) - 1

f(x) = 2 - 1

f(x) = 1

-

f(x) = 2x² - 1, x = 0

f(x) = 2(0)² - 1

f(x) = -1

-

f(x) = 2x² - 1, x = 1

f(x) = 2(1)² - 1

f(x) = 2(1) - 1

f(x) = 2 - 1

f(x) = 1

If two range values are listed, then you only need to write them once.

Therefore, the range of f(x) given the domain d = {-1,0,1} is r = {-1, 1}.

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

the answer for this question is : 360°

step-by-step explanation:

we have been given a triangle   we need to tell which one of these is a triangle below to map onto itself.

360° rotation is the complete rotation which eventually leads to the same figure which is given.

the correct will be, that is 360° that will cause the triangle below to map onto itself.