Use the Intermediate Value Theorem to prove that f(x) = x
3 − 9x + 5 has a real zero in each
of the following intervals: [−4, −3], [0, 1] and [2, 3].

The Intermediate Value Theorem (IVT) states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one real number \( c \) in the interval \((a, b)\) such that \( f(c) = 0 \). To use this theorem, we need to verify the continuity of the function and check the signs of \( f(a) \) and \( f(b) \) for each interval.

Given the function:

\[
f(x) = x^3 - 9x + 5
\]

Let's evaluate \( f(x) \) at the endpoints of each interval.

### 1. Interval \([-4, -3]\)

Evaluate \( f(-4) \) and \( f(-3) \):

\[
f(-4) = (-4)^3 - 9(-4) + 5 = -64 + 36 + 5 = -23
\]

\[
f(-3) = (-3)^3 - 9(-3) + 5 = -27 + 27 + 5 = 5
\]

Since \( f(-4) = -23 \) and \( f(-3) = 5 \), \( f(x) \) changes signs between \(-4\) and \(-3\). Therefore, by the Intermediate Value Theorem, there is at least one zero of \( f(x) \) in the interval \([-4, -3]\).

### 2. Interval \([0, 1]\)

Evaluate \( f(0) \) and \( f(1) \):

\[
f(0) = 0^3 - 9(0) + 5 = 5
\]

\[
f(1) = 1^3 - 9(1) + 5 = 1 - 9 + 5 = -3
\]

Since \( f(0) = 5 \) and \( f(1) = -3 \), \( f(x) \) changes signs between \(0\) and \(1\). Therefore, by the Intermediate Value Theorem, there is at least one zero of \( f(x) \) in the interval \([0, 1]\).

### 3. Interval \([2, 3]\)

Evaluate \( f(2) \) and \( f(3) \):

\[
f(2) = 2^3 - 9(2) + 5 = 8 - 18 + 5 = -5
\]

\[
f(3) = 3^3 - 9(3) + 5 = 27 - 27 + 5 = 5
\]

Since \( f(2) = -5 \) and \( f(3) = 5 \), \( f(x) \) changes signs between \(2\) and \(3\). Therefore, by the Intermediate Value Theorem, there is at least one zero of \( f(x) \) in the interval \([2, 3]\).

### Conclusion

By applying the Intermediate Value Theorem to each interval, we have shown that \( f(x) = x^3 - 9x + 5 \) has at least one real zero in each of the intervals \([-4, -3]\), \([0, 1]\), and \([2, 3]\).

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### Related Questions:

1. How do you determine if a function is continuous on a given interval?
2. What are other methods to find zeros of a polynomial function?
3. Can the Intermediate Value Theorem be used if the function is not continuous?
4. How does the IVT differ from other theorems like the Mean Value Theorem?
5. What are some real-world applications of the Intermediate Value Theorem?

### Tip:
Always check the continuity of the function first when applying the Intermediate Value Theorem.

Use the Intermediate Value Theorem to prove that f(x) = x^3 − 9x + 5 has a real zero in each of the following intervals: [−4, −3], [0, 1] and [2, 3].

Use the Intermediate Value Theorem to prove that the polynomial function f(x) = x^3 − 9x + 5 has real zeros in the intervals [−4, −3], [0, 1], and [2, 3]. This solution covers the continuity of functions and applies the IVT, suitable for students in Grades 10-12.

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