Identify square root of 2 as either rational or irrational, and approximate to the tenths place.
2 answers:
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Answer:
-4 4 4.5 8 20.5 32
Step-by-step explanation:
f g(x) = f(g(x))
Given, f(x) = 2x²
and g(x) = x - 2
Now f(g(x)) = f(x - 2) = 2(x - 2)²
We know that (a - b)² = a² - b² + 2ab
Using this we expand f(g(x)). We get:
f(g(x)) = 2{x² - 4x + 4}
Similarly, g(f(x)) = g(2x²) = 2x² - 2
Now, f(g(-2)) = 2[(-2)² - 4(-2) + 4] = 2(16) = 32.
Also, g(f(-2)) = 2[(-2)² - 2] = 2(2) = 4.
f(g(3.5)) = 2{(3.5)² -4(3.5) + 4} = 2[12.25 - 14 + 4] = 2(2.25) = 4.5.
g(f(3.5)) = 2{(3.5)² -2} = 2{12.25 - 2} = 2(10.25) = 20.5.
f(g(0)) = 2{0 - 4(0) + 4} = 2(4) = 8.
g(f(0)) = 2{0 - 2} = 2(-2) = -4.
Arranging them in ascending order, we get:
-4 4 4.5 8 20.5 32 would be the sequence.