Chapter 2 | Limits
189
Figure 2.38 There is a number c ∈ ⎡
⎤ ⎦ that satisfies
⎣ a , b
f ( c ) = z .
Example 2.36 Application of the Intermediate Value Theorem
Show that f ( x ) = x −cos x has at least one zero.
Solution Since f ( x ) = x −cos x is continuous over (−∞, +∞), it is continuous over any closed interval of the form ⎡ ⎣ a , b ⎤ ⎦ . If you can find an interval ⎡ ⎣ a , b ⎤ ⎦ such that f ( a ) and f ( b ) have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number c in ( a , b ) that satisfies f ( c ) =0. Note that f (0)=0−cos(0)=−1<0 and f ⎛ ⎝ π 2 ⎞ ⎠ = π 2 −cos π 2 = π 2 >0. Using the Intermediate Value Theorem, we can see that there must be a real number c in [0, π /2] that satisfies f ( c ) =0. Therefore, f ( x ) = x −cos x has at least one zero.
Example 2.37 When Can You Apply the Intermediate Value Theorem?
If f ( x ) is continuous over [0, 2], f (0) >0 and f (2) >0, can we use the Intermediate Value Theorem to conclude that f ( x ) has no zeros in the interval ⎡ ⎣ 0, 2]? Explain.
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