How do you use the Intermediate Value Theorem to show that the polynomial function # f(x) = (x^3)/2 - 4x + 1/(x)# has a root in the interval [-3, -1]?
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To use the Intermediate Value Theorem to show that the polynomial function ( f(x) = \frac{x^3}{2} - 4x + \frac{1}{x} ) has a root in the interval ([-3, -1]), we need to demonstrate that the function changes sign within that interval.
First, evaluate ( f(-3) ) and ( f(-1) ).
( f(-3) = \frac{(-3)^3}{2} - 4(-3) + \frac{1}{-3} = \frac{-27}{2} + 12 - \frac{1}{3} = -\frac{45}{6} + \frac{72}{6} - \frac{2}{6} = \frac{25}{6} > 0 )
( f(-1) = \frac{(-1)^3}{2} - 4(-1) + \frac{1}{-1} = -\frac{1}{2} + 4 - 1 = \frac{7}{2} > 0 )
Since ( f(-3) > 0 ) and ( f(-1) > 0 ), and ( f(x) ) is a continuous function, by the Intermediate Value Theorem, there exists at least one root of ( f(x) ) in the interval ([-3, -1]).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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