Given the function f (x) = x + 6x³ - x² -30x +4, use the Intermediate Value Theorem to decide which of the following intervals contains at least one zero. Select all that apply

To use the Intermediate Value Theorem to determine which intervals contain at least one zero, we need to find the signs of the function at the endpoints of each interval.

The function f(x) = x + 6x³ - x² - 30x + 4 is a polynomial function.

Let's evaluate f(x) at the endpoints of each interval and determine the signs:

1. Interval (-∞, -4):
The sign of f(-∞) = -∞ + 6(-∞)³ - (-∞)² - 30(-∞) + 4 = -∞ - ∞ + ∞ + ∞ + 4 = -∞
The sign of f(-4) = -4 + 6(-4)³ - (-4)² - 30(-4) + 4 = -4 - 384 - 16 + 120 + 4 = -280

2. Interval (-4, -2):
The sign of f(-4) = -280 (from the previous step)
The sign of f(-2) = -2 + 6(-2)³ - (-2)² - 30(-2) + 4 = -2 - 48 - 4 + 60 + 4 = 10

3. Interval (-2, 0):
The sign of f(-2) = 10 (from the previous step)
The sign of f(0) = 0 + 6(0)³ - (0)² - 30(0) + 4 = 4

4. Interval (0, 4):
The sign of f(0) = 4 (from the previous step)
The sign of f(4) = 4 + 6(4)³ - (4)² - 30(4) + 4 = 4 + 384 - 16 - 120 + 4 = 256

5. Interval (4, ∞):
The sign of f(4) = 256 (from the previous step)
The sign of f(∞) = ∞ + 6(∞)³ - (∞)² - 30(∞) + 4 = ∞ + ∞ - ∞ - ∞ + 4 = ∞

Based on the signs we found, the intervals that have a change in sign and therefore contain at least one zero are:

- Interval (-4, -2)
- Interval (0, 4)

So, the answer is:

(-4, -2), (0, 4)

To use the Intermediate Value Theorem, we need to find the values of f(x) for the endpoints of the intervals.

1. Interval [-4, -2]:
f(-4) = -4 + 6(-4)^3 - (-4)^2 - 30(-4) + 4 = -4 + 6(-64) - 16 + 120 + 4 = -4 - 384 - 16 + 120 + 4 = -280
f(-2) = -2 + 6(-2)^3 - (-2)^2 - 30(-2) + 4 = -2 + 6(-8) - 4 - 60 + 4 = -2 - 48 - 4 - 60 + 4 = -114

2. Interval [-2, 0]:
f(-2) = -2 + 6(-2)^3 - (-2)^2 - 30(-2) + 4 = -2 + 6(-8) - 4 - 60 + 4 = -2 - 48 - 4 - 60 + 4 = -114
f(0) = 0 + 6(0)^3 - (0)^2 - 30(0) + 4 = 0 + 6(0) - 0 - 0 + 4 = 0 + 0 - 0 + 4 = 4

3. Interval [0, 2]:
f(0) = 0 + 6(0)^3 - (0)^2 - 30(0) + 4 = 0 + 6(0) - 0 - 0 + 4 = 0 + 0 - 0 + 4 = 4
f(2) = 2 + 6(2)^3 - (2)^2 - 30(2) + 4 = 2 + 6(8) - 4 - 60 + 4 = 2 + 48 - 4 - 60 + 4 = -10

4. Interval [2, 4]:
f(2) = 2 + 6(2)^3 - (2)^2 - 30(2) + 4 = 2 + 6(8) - 4 - 60 + 4 = 2 + 48 - 4 - 60 + 4 = -10
f(4) = 4 + 6(4)^3 - (4)^2 - 30(4) + 4 = 4 + 6(64) - 16 - 120 + 4 = 4 + 384 - 16 - 120 + 4 = 256

From the calculations above, we can see that the function f(x) changes sign in the intervals [-4, -2] and [2, 4]. Therefore, these intervals contain at least one zero.

To decide which intervals contain at least one zero for the given function f(x) = x + 6x³ - x² - 30x + 4, we can use the Intermediate Value Theorem.

The Intermediate Value Theorem states that if f(x) is a continuous function on the closed interval [a, b], and if c is any number between f(a) and f(b), then there exists at least one number x = k in the open interval (a, b) such that f(k) = c.

To apply the Intermediate Value Theorem, we can evaluate the function at the endpoints of each interval and check if the sign changes.

Let's analyze the given intervals:

A) (-3, -2):
Evaluate f(-3) and f(-2) to determine if the signs change:
f(-3) = (-3) + 6(-3)³ - (-3)² - 30(-3) + 4 = -81 - 9 + 9 + 90 + 4 = 13
f(-2) = (-2) + 6(-2)³ - (-2)² - 30(-2) + 4 = -2 - 48 + 4 + 60 + 4 = 18

As the sign does change between f(-3) and f(-2) (f(-3) is positive and f(-2) is negative), there is at least one zero in the interval (-3, -2).

B) (-2, -1):
Evaluate f(-2) and f(-1) to determine if the signs change:
f(-2) = (-2) + 6(-2)³ - (-2)² - 30(-2) + 4 = -2 - 48 + 4 + 60 + 4 = 18
f(-1) = (-1) + 6(-1)³ - (-1)² - 30(-1) + 4 = -1 - 6 + 1 + 30 + 4 = 28

As the sign does change between f(-2) and f(-1) (f(-2) is positive and f(-1) is negative), there is at least one zero in the interval (-2, -1).

C) (0, 1):
Evaluate f(0) and f(1) to determine if the signs change:
f(0) = (0) + 6(0)³ - (0)² - 30(0) + 4 = 4
f(1) = (1) + 6(1)³ - (1)² - 30(1) + 4 = 1 + 6 - 1 - 30 + 4 = -20

As the sign does not change between f(0) and f(1) (both are positive), there may not be any zeros in the interval (0, 1).

D) (1, 2):
Evaluate f(1) and f(2) to determine if the signs change:
f(1) = (1) + 6(1)³ - (1)² - 30(1) + 4 = 1 + 6 - 1 - 30 + 4 = -20
f(2) = (2) + 6(2)³ - (2)² - 30(2) + 4 = 2 + 48 - 4 - 60 + 4 = -10

As the sign does not change between f(1) and f(2) (both are negative), there may not be any zeros in the interval (1, 2).

Based on the Intermediate Value Theorem, the intervals that contain at least one zero are (-3, -2) and (-2, -1).

use ± in the answers

I apologize for the confusion. Let's revise the answer using the ± symbol.

Based on the Intermediate Value Theorem, the intervals that contain at least one zero are:

(-3, -2)
(-2, -1)

So, the answer is: ± (-3, -2), ± (-2, -1)

options

±1,±2,±1/2,±1/7,±1/14,±2/7
±1,±2,±7,±14,±7/2
±1,±2,±7,±14
±1,±2,±1/7,±1/14

To determine which intervals contain at least one zero, we can use the given options.

The intervals that contain at least one zero are:

±1, ±2, ±1/7, ±1/14

So, the answer is: ±1, ±2, ±1/7, ±1/14.

Given the function f(x)=x^4+6x^3−x^2−30x+4, use the Intermediate Value Theorem to decide which of the following intervals contain at least one zero

To use the Intermediate Value Theorem to decide which intervals contain at least one zero for the function f(x) = x^4 + 6x^3 − x^2 − 30x + 4, we need to evaluate the function at the endpoints of each interval and determine if the signs of f(x) change.

1) Interval (-3, -2):
Evaluate f(-3) and f(-2) to determine if the signs change:
f(-3) = (-3)^4 + 6(-3)^3 − (-3)^2 − 30(-3) + 4 = 81 - 162 - 9 + 90 + 4 = 4
f(-2) = (-2)^4 + 6(-2)^3 − (-2)^2 − 30(-2) + 4 = 16 - 96 - 4 + 60 + 4 = -14

The sign changes between f(-3) and f(-2) (f(-3) is positive and f(-2) is negative), so the interval (-3, -2) contains at least one zero.

2) Interval (-2, -1):
Evaluate f(-2) and f(-1) to determine if the signs change:
f(-2) = (-2)^4 + 6(-2)^3 − (-2)^2 − 30(-2) + 4 = 16 - 96 - 4 + 60 + 4 = -14
f(-1) = (-1)^4 + 6(-1)^3 − (-1)^2 − 30(-1) + 4 = 1 - 6 - 1 + 30 + 4 = 28

The sign changes between f(-2) and f(-1) (f(-2) is negative and f(-1) is positive), so the interval (-2, -1) contains at least one zero.

3) Interval (0, 1):
Evaluate f(0) and f(1) to determine if the signs change:
f(0) = (0)^4 + 6(0)^3 − (0)^2 − 30(0) + 4 = 0 - 0 - 0 + 0 + 4 = 4
f(1) = (1)^4 + 6(1)^3 − (1)^2 − 30(1) + 4 = 1 + 6 - 1 - 30 + 4 = -20

The sign changes between f(0) and f(1) (f(0) is positive and f(1) is negative), so the interval (0, 1) contains at least one zero.

4) Interval (1, 2):
Evaluate f(1) and f(2) to determine if the signs change:
f(1) = (1)^4 + 6(1)^3 − (1)^2 − 30(1) + 4 = 1 + 6 - 1 - 30 + 4 = -20
f(2) = (2)^4 + 6(2)^3 − (2)^2 − 30(2) + 4 = 16 + 48 - 4 - 60 + 4 = 4

The sign does not change between f(1) and f(2) (both are positive), so the interval (1, 2) may not contain any zeros.

Based on the Intermediate Value Theorem, the intervals that contain at least one zero are (-3, -2), (-2, -1), and (0, 1).

answer with ± like in the previous question

I apologize for the oversight. Here is the revised answer with ± symbols:

Based on the Intermediate Value Theorem, the intervals that contain at least one zero are:

± (-3, -2)
± (-2, -1)
± (0, 1)

So, the answer is: ± (-3, -2), ± (-2, -1), ± (0, 1).

(-2,-1) is not an option but the other two are right. only fours answers are allowed

I apologize for the mistake. Thank you for pointing that out.

Based on the Intermediate Value Theorem, the intervals that contain at least one zero are:

± (-3, -2)
± (0, 1)

So, the correct answer is: ± (-3, -2), ± (0, 1).

theres 2 more answers

I apologize for the confusion. Let's re-evaluate the intervals and find the two remaining answers.

We already determined that the intervals (-3, -2) and (0, 1) contain at least one zero.

Now let's evaluate the remaining intervals:

4) Interval (1, 2):
Evaluate f(1) and f(2) to determine if the signs change:
f(1) = (1)^4 + 6(1)^3 − (1)^2 − 30(1) + 4 = 1 + 6 - 1 - 30 + 4 = -20
f(2) = (2)^4 + 6(2)^3 − (2)^2 − 30(2) + 4 = 16 + 48 - 4 - 60 + 4 = 4

There is a sign change between f(1) and f(2) (f(1) is negative and f(2) is positive), so the interval (1, 2) contains at least one zero.

5) Interval (-4, -3):
Evaluate f(-4) and f(-3) to determine if the signs change:
f(-4) = (-4)^4 + 6(-4)^3 − (-4)^2 − 30(-4) + 4 = 256 - 384 - 16 + 120 + 4 = -20
f(-3) = (-3)^4 + 6(-3)^3 − (-3)^2 − 30(-3) + 4 = 81 - 162 - 9 + 90 + 4 = 4

There is a sign change between f(-4) and f(-3) (f(-4) is negative and f(-3) is positive), so the interval (-4, -3) contains at least one zero.

Based on the Intermediate Value Theorem, the intervals that contain at least one zero are:

± (-4, -3)
± (-3, -2)
± (0, 1)
± (1, 2)

So, the full answer is: ± (-4, -3), ± (-3, -2), ± (0, 1), ± (1, 2).

solve the rational equation indicate any extraneous solutions 5/x + 8/(x+3) =7

To solve the rational equation (5/x) + (8/(x+3)) = 7, we will first multiply through by the common denominator, which is x(x+3):

x(x+3) * (5/x) + x(x+3) * (8/(x+3)) = x(x+3) * 7

This simplifies to:

5(x+3) + 8x = 7x(x+3)

Distribute and combine like terms:

5x + 15 + 8x = 7x^2 + 21x

Combine like terms again:

13x + 15 = 7x² + 21x

Rearrange the terms to form a quadratic equation:

7x² + 8x + 15 = 0

Now we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. However, upon factoring or using the quadratic formula, we find no real solutions for x.

Hence, the equation 5/x + 8/(x+3) = 7 has no real solutions and no extraneous solutions.

wrong

Apologies for the mistake. Let's solve the rational equation correctly:

To solve the rational equation (5/x) + (8/(x+3)) = 7, we can begin by finding a common denominator for the two fractions, which is x(x+3).

Multiplying through by the common denominator, we have:

(x(x+3))(5/x) + (x(x+3))(

what is the solution

solve the rational equation indicate any extraneous solutions 5/x + 8/x+3=7