The graph of a rational function has a local minimum at (7,0). The complex number 4 + 2i is a zero of the function. What is the least possible degree of the function?
2
3
4
5
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. Select 4 answers
[-5,-4]
[-4,-3]
[-3,-2]
[-1,0]
[0,1]
[1,2]
For the first question, we know that a local minimum at (7,0) implies that the degree of the function must be at least 4. We also have one zero at 4 + 2i, which is a non-real complex number. Since complex zeros always occur in conjugate pairs, the least possible degree of the function would be 4.
For the second question, we can use the Intermediate Value Theorem to determine which intervals contain at least one zero. The theorem states that if a function is continuous on a closed interval [a, b] and takes on different signs at the endpoints, then it must have at least one zero in that interval.
To apply the theorem to the given function f(x) = x^4 + 6x^3 - x^2 - 30x + 4, we can evaluate the function at the endpoints of each interval and check if the signs are different.
For [-5, -4]:
f(-5) = (-5)^4 + 6(-5)^3 - (-5)^2 - 30(-5) + 4 = 109 + 750 + 25 + 150 + 4 = 1038 (positive)
f(-4) = (-4)^4 + 6(-4)^3 - (-4)^2 - 30(-4) + 4 = 256 - 384 - 16 + 120 + 4 = -20 (negative)
Since the signs are different, the interval [-5, -4] contains at least one zero.
Similarly, we can evaluate the function at the endpoints of each interval to determine if they contain at least one zero. The intervals that contain at least one zero are:
[-5, -4]
[-3, -2]
[-1, 0]
[0, 1]
So the correct answers are:
[-5, -4]
[-3, -2]
[-1, 0]
[0, 1]
The graph of a rational function with a local minimum at (7,0) means that the function approaches infinity as x approaches both positive and negative infinity. This indicates that the function is not a simple polynomial, but a rational function with a vertical asymptote.
Since the complex number 4 + 2i is a zero of the function, the minimum possible degree of the function is 3. This is because complex zeros occur in conjugate pairs, so there must be at least one more zero in addition to 4 + 2i.
To determine which intervals contain at least one zero of the function f(x), we can evaluate the function at the endpoints of each interval using the Intermediate Value Theorem:
For the interval [-5,-4]:
f(-5) = (-5)^4 + 6(-5)^3 - (-5)^2 - 30(-5) + 4 = 396
f(-4) = (-4)^4 + 6(-4)^3 - (-4)^2 - 30(-4) + 4 = -20
The function changes sign between -5 and -4, so there is at least one zero in this interval.
For the interval [-4,-3]:
f(-4) = -20
f(-3) = (-3)^4 + 6(-3)^3 - (-3)^2 - 30(-3) + 4 = -68
The function does not change sign between -4 and -3, so there is no zero in this interval.
For the interval [-3,-2]:
f(-3) = -68
f(-2) = (-2)^4 + 6(-2)^3 - (-2)^2 - 30(-2) + 4 = 76
The function changes sign between -3 and -2, so there is at least one zero in this interval.
For the interval [-1,0]:
f(-1) = (-1)^4 + 6(-1)^3 - (-1)^2 - 30(-1) + 4 = -18
f(0) = 4
The function changes sign between -1 and 0, so there is at least one zero in this interval.
For the interval [0,1]:
f(0) = 4
f(1) = (1)^4 + 6(1)^3 - (1)^2 - 30(1) + 4 = -20
The function does not change sign between 0 and 1, so there is no zero in this interval.
For the interval [1,2]:
f(1) = -20
f(2) = (2)^4 + 6(2)^3 - (2)^2 - 30(2) + 4 = 48
The function changes sign between 1 and 2, so there is at least one zero in this interval.
Therefore, the intervals that contain at least one zero of the function f(x) are:
[-5,-4],
[-3,-2],
[-1,0],
[1,2].
To find the least possible degree of the rational function, we need to consider the given information.
Firstly, we know that the graph of the function has a local minimum at (7,0). This means that the function must be increasing before reaching (7,0) and decreasing after it. Since the function has to increase before decreasing, it must have at least one factor of x - 7.
Secondly, we are given that the complex number 4 + 2i is a zero of the function. This means that the function must have a factor of (x - 4 - 2i). However, since complex roots always come in conjugate pairs, the function must also have the factor (x - 4 + 2i).
Combining these factors, we have:
(x - 7) * (x - 4 - 2i) * (x - 4 + 2i)
Multiplying out this expression, we get:
(x - 7) * ((x - 4)^2 - (2i)^2)
(x - 7) * ((x - 4)^2 - 4)
Expanding further, we have:
(x - 7) * (x^2 - 8x + 16 - 4)
Simplifying, we get:
(x - 7) * (x^2 - 8x + 12)
This polynomial is of degree 3. Therefore, the least possible degree of the rational function is 3.
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To use the Intermediate Value Theorem to decide which intervals contain at least one zero of the function f(x)=x^4+6x^3-x^2-30x+4, we need to check if the function changes sign between the given intervals.
We evaluate the function at the endpoints of each interval and observe the sign changes:
For the interval [-5, -4]:
f(-5) = (-5)^4 + 6(-5)^3 - (-5)^2 - 30(-5) + 4 = 119
f(-4) = (-4)^4 + 6(-4)^3 - (-4)^2 - 30(-4) + 4 = 220
Since f(-5) < 0 and f(-4) > 0, there is a sign change, indicating the presence of at least one zero between -5 and -4.
For the interval [-4, -3]:
f(-4) = 220
f(-3) = (-3)^4 + 6(-3)^3 - (-3)^2 - 30(-3) + 4 = -58
Since f(-4) > 0 and f(-3) < 0, there is another sign change, indicating the presence of at least one zero between -4 and -3.
For the interval [-3, -2]:
f(-3) = -58
f(-2) = (-2)^4 + 6(-2)^3 - (-2)^2 - 30(-2) + 4 = 24
There is no sign change between these endpoints, indicating that there is no zero in this interval.
For the interval [-1, 0]:
f(-1) = (-1)^4 + 6(-1)^3 - (-1)^2 - 30(-1) + 4 = 38
f(0) = (0)^4 + 6(0)^3 - (0)^2 - 30(0) + 4 = 4
Since f(-1) > 0 and f(0) > 0, there is no sign change in this interval, indicating that there is no zero.
Therefore, the intervals [-5, -4] and [-4, -3] contain at least one zero of the function f(x)=x^4+6x^3-x^2-30x+4.
Which of the following is the domain of the function f(x)=x^2-4/4x^3-7x^2-2x ?
(-∞,-2) U (-2,1/4) U (1/4, ∞)
(-∞, 1/4) U (1/4,0) U (0,∞)
(-∞,-1/4) U (-1/4,0) U (0,2) U (2,∞)
(-∞,-1/4) U (-1/4,0) U (0,∞)
To find the domain of the function f(x) = (x^2 - 4) / (4x^3 - 7x^2 - 2x), we need to determine the values of x for which the function is defined.
The function is defined for all values of x except where the denominator is equal to zero. This means we need to find the values of x that make 4x^3 - 7x^2 - 2x equal to zero.
We can factor out an x from the expression:
x(4x^2 - 7x - 2) = 0
To find the zeros of the expression 4x^2 - 7x - 2, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For this expression, a = 4, b = -7, and c = -2.
Plugging in these values into the quadratic formula:
x = (-(-7) ± √((-7)^2 - 4(4)(-2))) / (2(4))
x = (7 ± √(49 + 32)) / 8
x = (7 ± √81) / 8
x = (7 ± 9) / 8
We have two possible values for x:
x = (7 + 9) / 8 = 16/8 = 2
x = (7 - 9) / 8 = -2/8 = -1/4
Therefore, the function is not defined at x = -1/4 and x = 2.
The domain of the function is all real numbers except for x = -1/4 and x = 2.
This can be represented as:
(-∞, -1/4) U (-1/4, 2) U (2, ∞)
Therefore, the correct answer is (-∞, -1/4) U (-1/4, 2) U (2, ∞).
Construct a rational function that has a vertical asymptote at x = 3 and a removable discontinuity at x=-2 . Explain how you determined your answer.
To construct a rational function with a vertical asymptote at x = 3 and a removable discontinuity at x = -2, we can follow these steps:
1. Start with a basic rational function in the form f(x) = P(x) / Q(x), where P(x) and Q(x) are polynomials.
2. To create a vertical asymptote at x = 3, we need to include a factor of (x - 3) in the denominator Q(x), since a rational function has a vertical asymptote at x = c if and only if the denominator has a factor of (x - c).
3. To create a removable discontinuity at x = -2, we need to make the numerator P(x) have a factor of (x + 2) and the denominator Q(x) have a corresponding factor of (x + 2). This will result in a common factor of (x + 2) that will cancel out and create the removable discontinuity.
4. Choose the degree of the polynomials P(x) and Q(x) to control the behavior of the function. The degree of Q(x) should be at least one degree higher than the degree of P(x) to ensure the vertical asymptote at x = 3.
5. Finally, adjust the coefficients of the polynomials to achieve the desired behavior and values.
One possible rational function that satisfies these conditions is:
f(x) = (x + 2) / ((x - 3)(x + 2))
In this function, the numerator is (x + 2), which has a factor of (x + 2) that will cancel out with the corresponding factor in the denominator. This creates a removable discontinuity at x = -2.
The denominator is (x - 3)(x + 2), which has a factor of (x - 3). This results in a vertical asymptote at x = 3. The degree of the denominator is one degree higher than the degree of the numerator, ensuring the presence of the vertical asymptote.
By adjusting the coefficients of the numerator and denominator, we can further control the behavior and values of the function.
Note: While this is one possible solution, there are other variations that could also satisfy the given conditions.
Find the equation of the horizontal asymptote of the function f(x)=x^3+1/5x^2-3x^3.
y=5/3
y=-1/3
y=0
y=1/5
To find the equation of the horizontal asymptote of the function f(x) = (x^3 + 1) / (5x^2 - 3x^3), we can examine the behavior of the function as x approaches positive and negative infinity.
As x approaches positive infinity, both the numerator and denominator of the function tend to infinity. In this case, the degree of the numerator (3) is greater than the degree of the denominator (2). Therefore, the numerator grows faster than the denominator, and the function approaches positive infinity as x approaches positive infinity. This means that the horizontal asymptote is y = +∞.
Similarly, as x approaches negative infinity, both the numerator and denominator of the function tend to infinity. Again, the degree of the numerator (3) is greater than the degree of the denominator (2). Therefore, the numerator grows faster than the denominator, and the function approaches negative infinity as x approaches negative infinity. This means that the horizontal asymptote is y = -∞.
Since the function has vertical asymptotes at x = 3 and y = -2, there is no horizontal asymptote at y = 0.
Thus, the correct answer is y = +∞ and y = -∞, as the function does not approach a specific value as x approaches infinity.
What is the equation of the slant asymptote of the graph of the function f(x)= 2x^2 -7x-8/x-4?
y=2x+2
y=x+3
y=2x+1
y=2x-15