The function h(x) is continuous on the interval [0,6]. If h(0)=–9 and h(6)=11, can you conclude that h(x) is ever equal to 1?
Question
The function h(x) is continuous on the interval [0,6]. If h(0)=–9 and h(6)=11, can you conclude that h(x) is ever equal to 1?
Solution
Yes, we can conclude that h(x) is ever equal to 1. This is due to the Intermediate Value Theorem (IVT). The IVT states that if a function is continuous on a closed interval [a,b], then it takes on every value between f(a) and f(b) at some point within the interval.
Given that h(x) is continuous on the interval [0,6] and h(0) = -9 and h(6) = 11, the function must take on all values between -9 and 11 (inclusive) at some point in the interval. Since 1 is between -9 and 11, there must be at least one value c in the interval [0,6] such that h(c) = 1.
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