Verify these answers :]
Would be thankies~
1. Determine the intervals in which the reciprocal function of f(x)= x^2+1 is increasing.
a) (0,∞)
b) (-∞, 0)
c) (-∞,∞)
d) (1,∞)
Answer: D
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2. Determine the point(s) where f(x)=2x^2-1 and its reciprocal function intersect.
a) (1,1)
b) (-1,1)
c) (1,1) and (-1,1)
d) (1,1), (-1,1) and (0,-1)
Answer:D
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3. Identify the vertical and horizontal asympototes of f(x)=x-4/2x+1
a)vertical x=4, horizontal:y=-1/2
b)vertical x=4, horizontal:y=1/2
c)vertical x=-1/2, horizontal:y=-1/2
d)vertical x=1/2, horizontal:y=-1/2
Answer: A
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4. State the equation of the rational function that meets these conditions:
-vertical asympotote at x=2
-Horizontal asympotote at y=1
-increases on each interval of its domain
-x-intecept is (3,0)
a) y=x+3/x+2
b) y=x-3/x+2
c) y=x+3/x-2
d) y=x-3/x-2
Answer: A
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5. state the equation of f(x) is D= [XER|x≠2/3] and the y-intercept is (0,1/2).
a) f(x)=2x+1/3x-2
b) f(x)=x-1/3x-2
c) f(x)=x+1/3x+2
d) f(x)=2x+1/3x+2
Answer is B?
1.
y = x^2+1
This is always positive, with its minimum value at x=0, where y=1. The range is [1,∞0, but that's not where it is increasing.
Now, the question asks about the reciprocal function z = 1/y
z = 1/(x^2+1)
Where y increases, z decreases, and vice versa. So, z is increasing on (-∞, 0), answer B
2. OK
3. None of the above. Would be C, but the horizontal is y = 1/2. I assume you wanted (x-4)/(2x+1)
4. vertical at x=2 means C or D
horizontal at y=1 keeps C or D
intercept at (3,0) means D
increasing keeps D
so, D.
5. D = x not 2/3 means A or B
y-intercept at (0,1/2) means B.
So, B.
1. Well, if you ask me, I'd say the reciprocal function of f(x) = x^2+1 is just begging to increase its coolness factor on the interval (1, ∞). So, the answer is definitely not D, it's A.
2. Ah, the intersection of f(x)=2x^2-1 and its reciprocal function. It's like two friends crossing paths in a parallel universe. They meet at the points (1, 1), (-1, 1), and (0, -1). So, the answer is not just D, it's D for "Deliciously overlapping points."
3. Vertical and horizontal asymptotes in the house! The vertical one throws a party at x = 4, while the horizontal one pulls out some smooth moves at y = -1/2. So, the answer is not A, it's A for "Asymptotes in da house!"
4. Oh, so you want the equation of a rational function that's feeling pretty confident. It winks at the vertical asymptote x = 2 and nods at the horizontal asymptote y = 1. To top it off, it high-fives the x-axis at (3,0). It's definitely NOT B, it's A, because A brings its A-game to every condition.
5. Ah, the equation of f(x) with some domain restrictions and a sassy y-intercept. It's like a math party with a bouncer at x = 2/3. As for the y-intercept, the function rocks it at (0, 1/2). So, the answer is not B, it's B for "Bringin' the domain and intercept vibes."
1. To determine the intervals in which the reciprocal function of f(x) = x^2 + 1 is increasing, we need to find where the derivative of the reciprocal function is positive.
Reciprocal function of f(x) = x^2 + 1: g(x) = 1 / (x^2 + 1)
To find the derivative of g(x), we can use the quotient rule:
g'(x) = (1 * (2x)) / ((x^2 + 1)^2) - (0) / (x^2 + 1)
g'(x) = 2x / (x^2 + 1)^2
We need to find where g'(x) > 0.
Since the denominator is always positive, we can just focus on the numerator.
2x > 0
x > 0
So the reciprocal function is increasing for x > 0.
Therefore, answer (a) (0,∞) is correct.
2. To find the point(s) where f(x) = 2x^2 - 1 and its reciprocal function intersect, we need to set the two functions equal to each other and solve for x.
f(x) = 2x^2 - 1
g(x) = 1 / f(x)
Setting the two functions equal to each other:
1 / f(x) = 2x^2 - 1
Now, we can solve for x.
1 = (2x^2 - 1) * f(x)
1 = (2x^2 - 1) * (1 / f(x))
1 = 2x^2 - 1
2x^2 = 2
x^2 = 1
x = ±1
So the functions intersect at x = 1 and x = -1.
Therefore, answer (c) (1,1) and (-1,1) is correct.
3. To identify the vertical and horizontal asymptotes of f(x) = (x - 4) / (2x + 1), we need to analyze the behavior of the function as x approaches infinity or negative infinity.
Vertical asymptote:
The vertical asymptote occurs when the denominator becomes zero. In this case, when 2x + 1 = 0.
2x = -1
x = -1/2
So the vertical asymptote is x = -1/2.
Horizontal asymptote:
To find the horizontal asymptote, we compare the degrees of the numerator and denominator.
The degree of the numerator is 1, and the degree of the denominator is also 1. When the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients.
The leading coefficients are 1 and 2, so the horizontal asymptote is y = 1/2.
Therefore, answer (a) vertical x = 4, horizontal y = -1/2 is incorrect.
The correct answer is (a) vertical x = -1/2, horizontal y = 1/2.
4. To state the equation of the rational function that meets the given conditions, we need to consider the vertical and horizontal asymptotes, the increasing intervals, and the x-intercept.
Vertical asymptote at x = 2:
The denominator of the rational function should be (x - 2).
Horizontal asymptote at y = 1:
The leading coefficient of the numerator should be equal to the leading coefficient of the denominator.
Increases on each interval of its domain:
To ensure the function increases on each interval of its domain, we'll use a positive coefficient for the x term in the numerator.
x-intercept at (3, 0):
The numerator should be (x - 3) to have the x-intercept at (3, 0).
Putting all these together, we get:
y = (x - 3) / (x - 2)
Therefore, answer (a) y = x + 3 / x + 2 is incorrect.
The correct answer is (b) y = x - 3 / x + 2.
5. To state the equation of f(x) with a domain D = [xER|x≠2/3] and a y-intercept at (0, 1/2), we can use the point-slope form of a linear equation.
Since the y-intercept is (0, 1/2), we have the point (0, 1/2) on the graph.
Using the point-slope form with (0, 1/2) and the slope 2:
y - 1/2 = 2(x - 0)
Simplifying:
y - 1/2 = 2x
Now we can rewrite the equation in the desired form:
y = 2x + 1/2
Therefore, the correct answer is (c) f(x) = x + 1 / 3x + 2.
1. To determine the intervals in which the reciprocal function of f(x) = x^2 + 1 is increasing, we need to find the intervals where the original function f(x) is decreasing.
First, find the derivative of f(x): f'(x) = 2x.
Now, set f'(x) = 0 and solve for x to find the critical points. In this case, there are no critical points because f'(x) = 2x is never equal to zero.
Since f'(x) is always positive (because 2x is always positive except at x = 0), the original function f(x) is always increasing. Therefore, the reciprocal function is always increasing as well.
The correct answer is c) (-∞,∞).
2. To determine the point(s) where f(x) = 2x^2 - 1 and its reciprocal function intersect, we need to find the x-values where the two functions are equal.
Set f(x) equal to its reciprocal 1/f(x) and solve for x:
2x^2 - 1 = 1/(2x^2 - 1)
Multiply both sides by (2x^2 - 1) to get rid of the denominator:
(2x^2 - 1)(2x^2 - 1) = 1
Expand and simplify:
4x^4 - 4x^2 + 1 - 1 = 0
4x^4 - 4x^2 = 0
Factor out 4x^2:
4x^2(x^2 - 1) = 0
Set each factor equal to zero:
4x^2 = 0 --> x = 0 (multiplicity 2) (1st factor)
x^2 - 1 = 0 --> x = ±1 (2nd factor)
Therefore, the points where the two functions intersect are (0, 1), (-1, 1), and (1, 1).
The correct answer is d) (1,1), (-1,1), and (0,1).
3. To identify the vertical and horizontal asymptotes of f(x) = (x - 4) / (2x + 1), we need to analyze the behavior of the function as x approaches positive and negative infinity.
- Vertical asymptote:
Set the denominator equal to zero and solve for x:
2x + 1 = 0 --> x = -1/2
Therefore, there is a vertical asymptote at x = -1/2.
- Horizontal asymptote:
To find the horizontal asymptote, we compare the degrees of the numerator and denominator:
- The degree of the numerator is 1 (highest power of x is x^1).
- The degree of the denominator is also 1 (highest power of x is x^1).
Since the degrees are the same, divide the coefficient of the highest power of x in the numerator by the coefficient of the highest power of x in the denominator:
1 / 2 = 1/2
Therefore, there is a horizontal asymptote at y = 1/2.
The correct answer is a) vertical x = 4, horizontal y = 1/2.
4. To state the equation of the rational function meeting certain conditions, we can use the given information to construct the equation.
- Vertical asymptote at x = 2:
This gives us a factor of (x - 2) in the denominator.
- Horizontal asymptote at y = 1:
Since the degrees of the numerator and denominator are the same, the horizontal asymptote is determined by the ratio of the leading coefficients. The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is 1. Therefore, y = 1 is the horizontal asymptote.
- Increases on each interval of its domain:
To ensure the function increases on each interval, the numerator's leading coefficient (1) must be positive.
- x-intercept at (3, 0):
This gives us a factor of (x - 3) in the numerator.
Based on this information, the equation of the rational function is:
f(x) = (x - 3) / (x - 2)
The correct answer is a) y = (x + 3) / (x + 2).
5. To state the equation of f(x) given a domain and a y-intercept, use the information provided.
- Domain D = [xER | x ≠ 2/3]:
This means that for any value of x, except x = 2/3, the function is defined.
- y-intercept at (0, 1/2):
This means that when x = 0, the function has a y-value of 1/2.
To construct the equation, we can use the standard form for a linear function, f(x) = mx + b, where m represents the slope and b represents the y-intercept.
- The slope (m) can be found using the y-intercept and an arbitrary point on the domain. Let's use the point (2, 0) since it is one of the points excluded from the domain:
m = (0 - 1/2) / (2 - 0)
m = -1/4
- Now we can use the slope (m = -1/4) and the y-intercept (b = 1/2) to form the equation:
f(x) = -1/4x + 1/2
The correct answer is c) f(x) = x + 1 / (3x + 2).