plx need help calc?

Question

let f be the function given by f(x)= x/root x^2-4
a) find the domain of f
b) write an equation for each vertival asymptote to the graph of f
c) write an equation for each horizontal asyptote to the graph of f
d) find f'(x)

Answer 1

a) The domain is all real numbers except x=2 and x=-2
b) x=2 and x=-2
c) y=1 and y=-1
d) f'(x)=[1-x^2(x^2-4)^(1/2)] / (x^2-4)^(3/2)

For horizontal asymptotes, the poster below me seems to have forgotten that on the negative side of the domain, there will be a horizontal asymptote at y = -1, not y = 1. As x -> negative infinity, the numerator gets arbitrarily negative while the denominator gets arbitrarily POSITIVE at the exact same rate. You can use L'Hospital's rule to prove that there will also be an asymptote at y = -1.

Haha. So much for us math people. I like the old adage that "math majors can't add."

My fixed domain:

a) D:{ x | All real numbers, x^2>2 }

My derivative process:

f(x) = x/(x^2-4)^(1/2)
f'(x) = [(x^2-4)^(1/2) - x(1/2)(2x)(x^2-4)^(-1/2)] / (x^2-4)
That was where I messed up. I just dropped down the (1/2) when I did the chain rule isntead of subtracting 1 from 1/2.
f'(x) = [(x^2-4)^(-1/2)(x^2 - 4 -x^2)]/(x^2-4)

Thus my revised (your original) derivative:

d) f'(x) = -4/(x^2-4)^(3/2)

Answer 2

f(x) = x / sqrt(x^2 - 4)

a) To find the domain of f(x), keep in mind the restrictions of the square root and the restriction of fractions.

For square roots, the inside MUST be greater than or equal to 0. For fractions, the denominator cannot equal 0. Combining these two facts, it follows that x^2 - 4 must be strictly greater than 0. All we have to do is solve that inequality.

x^2 - 4 > 0
(x - 2) (x + 2) > 0

Our critical values are x = {-2, 2}. What we want to do now is test values AROUND -2 and 2, to determine the intervals where it is positive, and where it is negative.

We first test a value less than -2. Test -10. Then
(x - 2) (x + 2) = (-10 - 2) (-10 + 2) = (negative) x (negative) = positive. Therefore, x^2 - 4 > 0 when x < -2.

Test a value between -2 and 2; test 0. Then
(x - 2) (x + 2) = (-2)(2) = (negative)(positive) = negative.
Therefore, x^2 - 4 < 0 when -2 < x < 2.

Test a value greater than 2. Test 10. Then
(x - 2) (x + 2) = (8)(12) = positive.
Therefore, x^2 - 4 > 0 when x > 2.

That means our inequality holds true when x > 2 or x < -2. That is our domain.

If we want our domain in interval notation, it is:
(-infinity, -2) U (2, infinity)

b) To find the vertical asymptotes, all we have to do is equate the denominator to 0, because these are the values x CANNOT be.

sqrt(x^2 - 4) = 0
x^2 - 4 = 0
(x - 2)(x + 2) = 0
Therefore, x = {-2, 2}
So our vertical asymptotes are x = -2 and x = 2.

c) To find the horizontal asymptotes, we have to solve the limit of f(x) as x approaches infinity, and the limit of f(x) as x approaches -infinity.

lim [x/sqrt(x^2 - 4)] =
x -> infinity

lim [x/sqrt(x^2(1 - 4/x^2))] =
x -> infinity

Pulling the x^2 out of the square root, x^2 = |x|

lim [x/|x|*sqrt(1 - 4/x^2)] =
x -> infinity

But |x| = x for positive values of x, so

lim [x/x*sqrt(1 - 4/x^2)] =
x -> infinity

Now we can cancel the x terms, which allows us to plug in infinity directly.

lim [1/sqrt(1 - 4/x^2)] = 1/sqrt(1 - 0) = 1/1 = 1
x -> infinity

Therefore, a horizontal asymptote would be at y = 1.

Now we solve the same thing, to -infinity. The steps are similar, but when we get to this point:

lim [x/|x|*sqrt(1 - 4/x^2)] =
x -> -infinity

|x| is equal to -x , since x is approaching negative infinity.

lim [x/-x*sqrt(1 - 4/x^2)] =
x -> -infinity

lim [(-1)/sqrt(1 - 4/x^2)] = (-1)/sqrt(1 - 0) = -1
x -> -infinity

So the second horizontal asymptote would be y = -1
The horizontal asymptotes are y = 1 and y = -1.

(d) To find f'(x), all we need to do is use the quotient rule.
Verbally, it goes "the derivative of the top times the bottom, minus the derivative of the bottom times the top, over the bottom squared."

A few things to note that d/dx sqrt(x) = 1/2sqrt(x). We're going to use this, along with the chain rule.

f'(x) = [(1)sqrt(x^2 - 4) - (1/[2sqrt(x^2 - 4)])(2x)(x)] / [x^2 - 4]
f'(x) = [sqrt(x^2 - 4) - (2x^2/[2sqrt(x^2 - 4)] / [x^2 - 4]

Multiplying numerator and denominator to get rid of all fractions by sqrt(x^2 - 4), some of the radicals go away, and we end up with

f'(x) = [ x^2 - 4 - 2x^2/2 ] / [x^2 - 4]^(3/2)
f'(x) = [ x^2 - 4 - x^2 ] / [x^2 - 4]^(3/2)
f'(x) = (-4) / [x^2 - 4]^(3/2)

Answer 3

f(x) = x / √(x^2 - 4)

(a) The domain of f is {x ∈ R: x^2 - 4 > 0} = (-infinity, -2) ∪ (2, infinity).

(b) For vertical asymptotes we need the denominator to be 0, so this can only be at 2 or -2. f(x) tends to -infinity as x goes to -2 from below and to +infinity as x goes to 2 from above, so x = 2 and x = -2 are the equations for the vertical asymptotes.

(c) For horizontal asymptotes we look at the limit of f as x goes to positive or negative infinity. If we divide top and bottom by x we get
f(x) = 1 / √(1 - 4/x^2), x > 2; -1 / √(1 - 4/x^2), x < -2

Obviously as x tends to positive or negative infinity, this tends to 1 or -1 respectively. So there are horizontal asymptotes with equations y = 1 and y = -1.

(d) Use the quotient rule:
f'(x) = [1.√(x^2-4) - x.(1/2)(x^2-4)^(-1/2).(2x)] / [x^2 - 4]
= (x^2-4)^(-1/2) [x^2 - 4 - x^2] / (x^2 - 4)
= -4 / (x^2 - 4)^(3/2)

Patrick, yes, I'd made an error which I realised and corrected after seeing your answer - note that your domain is still wrong as of this edit (as was mine initially. Shows what not being careful can do). I don't think I believe your derivative, either...

Answer 4

The domain is x < -2, x > 2
for -2 < x < 2 the function is complex

The vertical asymptotes are
x = - 2
x = 2

The horizontal asymptotes are
y = - 1
y = 1

f(x)= x(x^2 - 4)^-(1/2)
f'(x) = -2(1/2)x^2(x^2 - 4)^-(3/2) + (x^2 - 4)^-(1/2)
f'(x) = - x^2/(x^2 - 4)^(3/2) + (x^2 - 4)/(x^2 - 4)^(3/2)
f'(x) = - 4/(x^2 - 4)^(3/2)

Answer 5

a Domain: x>or equal to -2 and x b Vertical asymptotes at x=2 and x=-2
c Horizontal asymptote at y=0
d dy/dx=((x^2-4)^.5+x^2(x^2-4)^-.5)/(x^2-4)

Answer 6

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