194k views
1 vote
Use the Intermediate Value Theorem to show that

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

User Crazii
by
8.4k points

1 Answer

6 votes

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
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories