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zloy xaker [14]
3 years ago
10

,,,,,??,??,,,,,,,,,,,,,,,,

Mathematics
1 answer:
Andre45 [30]3 years ago
7 0

Answer:

  the one real zero is in the interval (-1, 0)

Step-by-step explanation:

Descartes' rule of signs tells you there are 0 or 2 positive real zeros. Changing the signs of the odd-degree terms and applying that rule again tells you there is one negative real zero. At the same time, those coefficients (-3, -5, -5, +7) have a negative sum, so you know ...

  f(-1) = -6

  f(0) = +7

so there is a zero in the interval (-1, 0).

__

You can try a few values between x=0 and x=10 to see what the function does in that part of the graph. You find ...

  f(1) = 10

  f(2) = 21

  f(3) = 58

So, it is safe to conclude that there are no real zeros for x > 0.

The only real zero of f(x) is in the interval (-1, 0).

_____

I like to use a graphing calculator for problems like this.

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