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AP Calc AB. IVT. Introduction. What do we know GENERALLY about the solutions for the functions below? f(x)= cos x -2 g(x) = h(x) =. Because the function is continuous, it must take on every y value between and. Intermediate Value Theorem.
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AP Calc AB IVT
Introduction What do we know GENERALLY about the solutions for the functions below? f(x)=cos x -2 g(x) = h(x) =
Because the function is continuous, it must take on every y value between and . Intermediate Value Theorem If a function is continuous between a and b, then it takes on every value between and .
The application of the IVT. If f(x) is continuous on a closed interval [a,b] and a<c<b, then there exists a value f(c) such that f(c) is between f(a) and f(b)
Example 1 Prove that there exists a value c such that f(c) = 0 for f(x) = for the interval [3,5]
Solution f(x) is continuous and f(3) = -6 and f(5) = 8 Therefore according to the IVT there exists a value c where -6 < f(c)< 8. Zero lies within that interval.
Example 2 Prove that there exists a value c such that f(c)=2 for f(x) = for the interval [-10,-3]
Solution f(x) is continuous and f(-10) = 8 and f(-3) = 1, therefore according to the IVT there exists a value c where 1<f(c)<8. 2 lies within this interval.
Root solution p(-1)=8 and p(2) = -19 p(x) is continuous and p(-1) = 8 and p(2) = -19, therefore according to the IVT there exists a value c between [-1, 2] where -19<p(c)<8. (or, you can simply say that there is sign change in the y values guaranteeing a value c in this interval where p(c)=0.) Note: we have not found c, just shown that there is at least ONE VALUE c.