Consider the function f(x)= 3
------.
x^2-25
a) Determine any restrictions on x.
b) State the domain and range.
c) State equation(s) for the asymptote(s).
d) Determine any x- and y-intercepts.
e) Sketch a graph of the function.
f) Describe the behaviour of the function as x approaches -5 and 5.
someone please kindly do this for me please. i beg you, please do all these for me please....
f(x) is not defined when the denominator is zero. So,
(a) x ≠ ±5
(b) domain is all reals except ±5
The denominator is minimum when x=0, so f(x) < -3/25 when -5<x<5
For |x| > 5 f(x) > 0, so the range is
(-∞,-3/25)∪(0,∞)
(c) vertical: x=±5
horizontal: y=0
(d) x: none
y: -3/25
(e) see wolframalpha.com
(f) x -> +∞ for |x| > 5
x -> -∞ for |x| < 5
Sure, I'd be happy to help you with this!
a) To determine any restrictions on x, we need to look at the denominator of the function. In this case, the denominator is x^2 - 25. We know that division by zero is undefined, so we need to find the values of x that make the denominator equal to zero. Equating the denominator to zero, we have x^2 - 25 = 0. Solving this equation, we get x = ±5. Therefore, the function is undefined when x = ±5.
b) The domain of a function is the set of all possible input values x, and the range is the set of all possible output values f(x). Since there are no other restrictions mentioned in the question, we can assume the domain is all real numbers except ±5. So, the domain is (-∞, -5) ∪ (-5, 5) ∪ (5, ∞). The range, on the other hand, includes all possible values of f(x). In this case, since the function is a rational function, it will have a horizontal asymptote at y = 0. So, the range is (-∞, 0) ∪ (0, ∞).
c) To find the equation(s) for the asymptote(s), we need to analyze the behavior of the function as x approaches positive and negative infinity. In this case, as x approaches positive infinity, f(x) approaches zero. Therefore, the horizontal asymptote is y = 0.
d) To determine the x-intercepts, we need to find the values of x that make f(x) equal to zero. In this case, f(x) = 0 when the numerator is equal to zero, which is when 3 = 0. However, there is no x-value that satisfies this condition, so the function has no x-intercepts.
To find the y-intercept, we need to substitute x = 0 into the function f(x). In this case, the y-intercept is when x = 0, resulting in f(0) = 3/(0^2 - 25) = 3/(-25) = -3/25. So, the y-intercept is (0, -3/25).
e) Now, let's sketch a graph of the function. Since we know the asymptote is y = 0, we can draw a horizontal line to represent it. The graph will approach the asymptote as x approaches positive or negative infinity. Also, since the function is undefined at x = ±5, there will be vertical asymptotes at x = 5 and x = -5. We can also plot the y-intercept at (0, -3/25).
f) As x approaches -5, f(x) will approach negative infinity. Similarly, as x approaches 5, f(x) will approach positive infinity. This is because the denominator x^2 - 25 approaches zero as x approaches ±5, causing the fraction to approach infinity.
I hope this explanation helps you understand how to approach these types of questions! If you have any further questions, feel free to ask.