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 contains at least one zero.
The answers are below but i need help with the explanation please
[-5,-4]
[-4,-3]
[0,1]
[1,2]
an interval must contain at least one root if f(x) changes sign in the interval. In other words, it can't go from positive to negative without attaining the value zero somewhere in between.
So, now you just need to check the intervals.
f(-5) = 4
f(-4) = -20
So, somewhere in [-5,-4] f(x) must be zero.
To use the intermediate value theorem, we need to check if there is a change in the sign of the function within each interval. If there is a change in sign, it means the function must cross the x-axis and have at least one zero within that interval.
Let's evaluate the function at the lower and upper bound of each interval and see if there is a change in sign.
Interval [-5, -4]:
f(-5) = (-5)^4 + 6(-5)^3 - (-5)^2 - 30(-5) + 4 = 765
f(-4) = (-4)^4 + 6(-4)^3 - (-4)^2 - 30(-4) + 4 = 4
Since f(-5) and f(-4) have a change in sign (positive to negative), there must be at least one zero in the interval [-5, -4].
Interval [-4, -3]:
f(-4) = 4
f(-3) = (-3)^4 + 6(-3)^3 - (-3)^2 - 30(-3) + 4 = -28
Since f(-4) and f(-3) have a change in sign (positive to negative), there must be at least one zero in the interval [-4, -3].
Interval [0, 1]:
f(0) = (0)^4 + 6(0)^3 - (0)^2 - 30(0) + 4 = 4
f(1) = (1)^4 + 6(1)^3 - (1)^2 - 30(1) + 4 = -20
Since f(0) and f(1) have a change in sign (positive to negative), there must be at least one zero in the interval [0, 1].
Interval [1, 2]:
f(1) = -20
f(2) = (2)^4 + 6(2)^3 - (2)^2 - 30(2) + 4 = 52
Since f(1) and f(2) do not have a change in sign, there may not be any zeros in the interval [1, 2].
In conclusion, the intervals [-5, -4], [-4, -3], and [0, 1] contain at least one zero, while the interval [1, 2] may not contain any zero for the function f(x) = x^4 + 6x^3 - x^2 - 30x + 4.
To use the Intermediate Value Theorem (IVT), we need to check if the function f(x) crosses the x-axis (i.e., has a zero) within each of the given intervals.
1) The interval [-5, -4]:
- Evaluate f(-5) and f(-4).
- If f(-5) and f(-4) have opposite signs, then by the IVT, there exists at least one zero within the interval [-5, -4].
2) The interval [-4, -3]:
- Evaluate f(-4) and f(-3).
- If f(-4) and f(-3) have opposite signs, then by the IVT, there exists at least one zero within the interval [-4, -3].
3) The interval [0, 1]:
- Evaluate f(0) and f(1).
- If f(0) and f(1) have opposite signs, then by the IVT, there exists at least one zero within the interval [0, 1].
4) The interval [1, 2]:
- Evaluate f(1) and f(2).
- If f(1) and f(2) have opposite signs, then by the IVT, there exists at least one zero within the interval [1, 2].
By evaluating the function at the endpoints of each interval and considering the sign of f(x), we can determine which intervals satisfy the conditions of the IVT and contain at least one zero.