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Use the Intermediate Value Theorem to show that

f(x) = 5x3 + 3x2 + 5x + 4 has a real zero between - 1 and 0.

User Crazii
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Final answer:

According to the Intermediate Value Theorem, since f(x) = 5x^3 + 3x^2 + 5x + 4 changes sign from f(-1) = -3 (negative) to f(0) = 4 (positive), there must be at least one real zero between -1 and 0.

Step-by-step explanation:

To show that the function f(x) = 5x^3 + 3x^2 + 5x + 4 has a real zero between -1 and 0 using the Intermediate Value Theorem (IVT), we evaluate the function at the endpoints of the interval [-1, 0].

First, let's evaluate f at x = -1:


f(-1) = 5(-1)^3 + 3(-1)^2 + 5(-1) + 4
= -5 + 3 - 5 + 4
= -3

Now, let's evaluate f at x = 0:


f(0) = 5(0)^3 + 3(0)^2 + 5(0) + 4
= 0 + 0 + 0 + 4
= 4

According to the IVT, if a continuous function changes sign over an interval, there must be at least one root in that interval. In this case, f(-1) = -3 and f(0) = 4, which shows a change in sign from negative to positive. Thus, there must be at least one real zero between -1 and 0.

User Poff
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