Consider the following function f(x)=x^2/[x^2-9]
f(x) is increasing on the interval(s)
f(x) is decreasing on the interval(s)
f(x) has 2 vertical asymptotes x=
f(x) is concave up on the interval(s)
f(x) concave down on the interval(s)
I've been stuck on these parts, I answered every other question like this, but cannot figure out this one.
vertical asymptotes is easy because
x = 3 or -3 makes function undefined
where is slope + and where - ?
slope = ([x^2-9]2x - x^2[2x])/[x^2-9]^2
denominator is always + or 0
when is top + and -?
top is -18x
so slope = -18x/positive number
so
slope is + if x is + and - if x is -
do next derivative same way, when numerator is + that is a bottom and when - that is a top
To determine if a function is increasing or decreasing on a given interval, we need to find the derivative of the function and examine its sign changes.
Let's start by finding the derivative of f(x) = x^2/(x^2 - 9). Using the quotient rule, we obtain:
f'(x) = [2x(x^2 - 9) - x^2(2x)] / (x^2 - 9)^2
Simplifying further:
f'(x) = (2x^3 - 18x - 2x^3) / (x^2 - 9)^2
f'(x) = (-18x) / (x^2 - 9)^2
Now, to determine where f(x) is increasing or decreasing, we need to analyze the sign changes of the derivative.
Let's first look for the critical points of f(x) where f'(x) might change sign by solving f'(x) = 0:
(-18x) / (x^2 - 9)^2 = 0
This occurs when x = 0. Therefore, x = 0 is a critical point.
Now, we need to consider the intervals between these critical points to determine where f(x) is increasing or decreasing.
1. For x < -3: Choose a value of x less than -3, for example, x = -4. Plug this value into f'(x) = (-18x) / (x^2 - 9)^2:
f'(-4) = (-18(-4)) / ((-4)^2 - 9)^2 = 72 / 25
Since f'(-4) > 0, f(x) is increasing on x < -3.
2. For -3 < x < 0: Using x = -2 as a test value, we find:
f'(-2) = (-18(-2)) / ((-2)^2 - 9)^2 = -36 / 49
Since f'(-2) < 0, f(x) is decreasing on -3 < x < 0.
3. For 0 < x < 3: Use x = 1 as a test value:
f'(1) = (-18(1)) / ((1)^2 - 9)^2 = -18 / 64
Since f'(1) < 0, f(x) is decreasing on 0 < x < 3.
4. For x > 3: Choose x = 4 as a test value:
f'(4) = (-18(4)) / ((4)^2 - 9)^2 = -72 / 175
Since f'(4) < 0, f(x) is decreasing on x > 3.
Therefore, we have:
- f(x) is increasing on x < -3,
- f(x) is decreasing on -3 < x < 0 and x > 3.
Now, let's move on to the remaining questions:
To find the vertical asymptotes of f(x), we examine where the denominator of f(x) becomes zero, that is, where x^2 - 9 = 0. Solving this equation gives:
x^2 - 9 = 0
(x - 3)(x + 3) = 0
x = 3 and x = -3
So, f(x) has two vertical asymptotes at x = 3 and x = -3.
To determine the concavity of f(x), we need to find the second derivative of f(x). Applying the quotient rule and simplifying, we get:
f''(x) = (18(x^4 - 9x^2 + 27)) / (x^2 - 9)^3
To find where f(x) is concave up or concave down, we need to examine the sign changes of f''(x).
For x < -3: Testing x = -4,
f''(-4) = (18((-4)^4 - 9(-4)^2 + 27)) / ((-4)^2 - 9)^3
= -3060 / 2025 = -12/5
Since f''(-4) < 0, f(x) is concave down on x < -3.
For -3 < x < 3: Testing x = 0,
f''(0) = (18(0^4 - 9(0)^2 + 27)) / (0^2 - 9)^3 = 0
Since f''(0) = 0, we cannot determine the concavity on this interval.
For x > 3: Testing x = 4,
f''(4) = (18((4)^4 - 9(4)^2 + 27)) / ((4)^2 - 9)^3
= 3060 / 2025 = 12/5
Since f''(4) > 0, f(x) is concave up on x > 3.
To summarize:
- f(x) is increasing on x < -3.
- f(x) is decreasing on -3 < x < 0 and x > 3.
- f(x) has two vertical asymptotes at x = 3 and x = -3.
- f(x) is concave down on x < -3.
- f(x) is concave up on x > 3.
I hope this helps you understand how to approach these types of questions!