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Nimfa-mama [501]
3 years ago
15

If f(x) = -(1/x^3) and x takes on successive values from -10 to 0.1 ,

Mathematics
1 answer:
prohojiy [21]3 years ago
3 0
Alright, this one is a little interesting... Let's perform some tests to figure out what is happening:
f(-10) = -(1/(-10)^3) = -(1/-1000) = 1/1000  (positive)
f(-5) = -(1/(-5)^3) = -(1/-125) = 1/125 (positive, bigger than the last one)
f(-1) = -(1/(-1)^3) = -(1/-1) = 1 (positive, bigger than the last one)
f(-0.1) = -(1/(-0.1)^3) = -(1/-0.001) = 1/0.001 = 1000 (positive, bigger than the last one)
f(0) = -(1/0^3) = undefined!
f(0.1) = -(1/(0.1)^3) = -(1/0.001) = -1/0.001 = -1000 (negative)
f(1) = -(1/1^3) = -(1/1) = -1 (negative, but bigger than last one)

It's a little confusing with the undefined part at x = 0. What I can say is this, it is increasing from -10 up to 0, something weird happens at 0 and it resets, and starts increasing from 0 up to 0.1. 

I guess A would be the best answer?

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