How do you use the Intermediate Value Theorem to show that the polynomial function # 2tanx - x - 1 = 0# has a zero in the interval [0, pi/4]?
and has a value
therefore
by the Intermediate Value Theorem
it has ha value
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To use the Intermediate Value Theorem to show that the polynomial function ( 2\tan(x) - x - 1 = 0 ) has a zero in the interval ([0, \frac{\pi}{4}]), we need to verify that the function changes sign over that interval.
Evaluate the function at the endpoints of the interval:
- At ( x = 0 ): ( 2\tan(0) - 0 - 1 = -1 )
- At ( x = \frac{\pi}{4} ): ( 2\tan\left(\frac{\pi}{4}\right) - \frac{\pi}{4} - 1 = 2 - \frac{\pi}{4} - 1 = \frac{7}{4} - \frac{\pi}{4} )
Since ( -1 < 0 < \frac{7}{4} - \frac{\pi}{4} ), and the function is continuous over the interval ([0, \frac{\pi}{4}]), by the Intermediate Value Theorem, there exists at least one value ( c ) in the interval ([0, \frac{\pi}{4}]) such that ( f(c) = 0 ), where ( f(x) = 2\tan(x) - x - 1 ). Therefore, the polynomial function has a zero in the interval ([0, \frac{\pi}{4}]).
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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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