given the function f(x)= x^4 -3x^3 -2x^2 + 5x + 1 use the intermediate value theorem to decide which of the following intervals contains at least one zero, select all that apply (4 answers)
a) [-2,-1]
b) [-1,0]
c) [0,1]
d) [1,2]
e) [2,3]
f) [3,4]
The answers are:
a) [-2, -1]
b) [-1, 0]
d) [1, 2]
f) [3, 4]
If you put the equation on a graph, just look at where the line crosses the x-axis and what numbers it's between when it crosses. Hope this helps! :)
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To determine which of the given intervals contains at least one zero using the intermediate value theorem, we need to evaluate the function at the endpoints of each interval and check if the sign of the function changes.
Let's evaluate the function at the endpoints of each interval:
a) [-2,-1]
f(-2) = (-2)^4 - 3(-2)^3 - 2(-2)^2 + 5(-2) + 1 = 11
f(-1) = (-1)^4 - 3(-1)^3 - 2(-1)^2 + 5(-1) + 1 = -1
Since the sign of the function changes from positive to negative between the endpoints of the interval, there is at least one zero in the interval [-2,-1].
b) [-1,0]
f(-1) = -1 (we have already evaluated this)
f(0) = 0^4 - 3(0)^3 - 2(0)^2 + 5(0) + 1 = 1
Since the sign of the function does not change between the endpoints of the interval, there is no zero in the interval [-1,0].
c) [0,1]
f(0) = 0 (we have already evaluated this)
f(1) = 1^4 - 3(1)^3 - 2(1)^2 + 5(1) + 1 = 2
Since the sign of the function does not change between the endpoints of the interval, there is no zero in the interval [0,1].
d) [1,2]
f(1) = 1 (we have already evaluated this)
f(2) = 2^4 - 3(2)^3 - 2(2)^2 + 5(2) + 1 = 25
Since the sign of the function changes from positive to negative between the endpoints of the interval, there is at least one zero in the interval [1,2].
e) [2,3]
f(2) = 25 (we have already evaluated this)
f(3) = 3^4 - 3(3)^3 - 2(3)^2 + 5(3) + 1 = -55
Since the sign of the function changes from positive to negative between the endpoints of the interval, there is at least one zero in the interval [2,3].
f) [3,4]
f(3) = -55 (we have already evaluated this)
f(4) = 4^4 - 3(4)^3 - 2(4)^2 + 5(4) + 1 = 221
Since the sign of the function does not change between the endpoints of the interval, there is no zero in the interval [3,4].
In conclusion, the intervals that contain at least one zero are:
a) [-2,-1]
d) [1,2]
e) [2,3]
To apply the Intermediate Value Theorem, we need to evaluate the function at the endpoints of each interval and check if the signs of the function values differ. If the signs differ, this means that the function must have at least one zero within that interval.
Let's evaluate the function f(x) = x^4 - 3x^3 - 2x^2 + 5x + 1 at the endpoints of each interval:
a) For the interval [-2, -1]:
f(-2) = (-2)^4 - 3(-2)^3 - 2(-2)^2 + 5(-2) + 1 = 16 + 24 - 8 - 10 + 1 = 23 > 0
f(-1) = (-1)^4 - 3(-1)^3 - 2(-1)^2 + 5(-1) + 1 = 1 + 3 - 2 - 5 + 1 = -2 < 0
The signs of f(-2) and f(-1) differ, so there is at least one zero in the interval [-2, -1].
b) For the interval [-1, 0]:
f(-1) = -2 < 0
f(0) = 1 > 0
The signs of f(-1) and f(0) differ, so there is at least one zero in the interval [-1, 0].
c) For the interval [0, 1]:
f(0) = 1 > 0
f(1) = 1 - 3 - 2 + 5 + 1 = 2 > 0
Both f(0) and f(1) are positive, so there is no zero in the interval [0, 1].
d) For the interval [1, 2]:
f(1) = 2 > 0
f(2) = 16 - 24 - 8 + 10 + 1 = -5 < 0
The signs of f(1) and f(2) differ, so there is at least one zero in the interval [1, 2].
e) For the interval [2, 3]:
f(2) = -5 < 0
f(3) = 81 - 81 - 18 + 15 + 1 = -2 < 0
Both f(2) and f(3) are negative, so there is no zero in the interval [2, 3].
f) For the interval [3, 4]:
f(3) = -2 < 0
f(4) = 256 - 432 - 32 + 20 + 1 = -187 < 0
Both f(3) and f(4) are negative, so there is no zero in the interval [3, 4].
Therefore, the intervals that contain at least one zero are: a) [-2, -1] and b) [-1, 0].
(a) f(-2) = 23 and f(-1) = -2
Since f(x) is continuous, there is no way to get from f(x) = 23 to f(x) = -2 without having f(x)=0 somewhere in the interval [-2,-1]
Now do the others in like wise
To confirm your answers graph the function and see where it crosses the x-axis.