i need answers ASAP: The relationship between x and y can be modeled by an inverse variation function eq Table What is the value
1 answer:
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Step-by-step explanation:
Q3
(a) If 0 < |x − 1| < δ, then |f(x) − 2| < 2
What this means is, how far can x stray from the x=1 line such that f(x) stays within 2 units of the y=2 line (0 < f(x) < 4).
If we move 2 units left of x=1, we get f(x) = 4.
If we move about 3.5 units right of x=1, we get f(x) = 4.
Therefore, δ can't be more than 2.
(b) If 0 < |x| < δ, then |f(x) − 3| < 1
What this means is, how far can x stray from the x=0 line such that f(x) stays within 1 unit of the y=3 line (2 < f(x) < 4).
If we move 1 unit left of x=0, we get f(x) = 4.
If we move 1 unts right of x=0, we get f(x) = 2.
Therefore, δ can't be more than 1.
Since f(x) isn't continuous within this domain, we can't conclude that the limit exists.
Q4
(a) Yes. If δ = 0.25, then 0.75 < x < 1.25, and f(x) > 200.
(b) No. f(1) = 300, so even if δ = 0, f(x) will be less than 400.
(c) Yes. If δ ≈ 0.1, then 0.9 < x < 1, and f(x) > 450.
