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).