Multiple Part Calc. Question?

Question

f(x)= (-3x)/(2-x) with that function how do you find...

domain
range
horizontal asymptote
vertical asymptote
slant asymptote
x-intercept
y-intercept
f(x) increases on
f(x) decreases on
relative maximum
relative minimum
f(x) is concave upward on
f(x) is concave downward on
point(s) of inflection


any part(s) of that you can answer along with work on how you got it will be very helpful

Answer 1

f(x) = (-3x) / (2 - x)

The first thing to do change the form of this question using synthetic long division. Without showing you the steps, you should get a quotient of 3 and a remainder of -6, so you can change f(x) to

f(x) = -6/(2 - x) + 3

To get the domain, all you have to recognize is that the denominator cannot equal 0. 2 - x = 0 when x = 2, so the domain would be all real numbers except 2. In interval notation,
Domain: (-infinity, 2) U (2, infinity)

To get the range, look at the quotient of the long division (that is the 3). The number which is NOT part of the fraction is what f(x) CANNOT be. Therefore the range is all real numbers except 3.
Range: (-infinity, 3) U (3, infinity)

The vertical asymptote compliments the domain of the function; since x cannot equal 2, your vertical asymptote will be x = 2.

Similarly, the horizontal asymptote will be y = 3.

There are no slant asymptotes, because the quotient of the long division determines whether there would be a slant asymptote. (for instance, if we had 1/(2 - x) + x, then our slant asymptote would be y = x).

To solve for the y-intercept, set x = 0, to obtain
f(0) = -6/(2 - 0) + 3 = -6/2 + 3 = -3 + 3 = 0.
Y-intercept is 0.

To solve for the x-intercept, set f(x) = 0. Then
0 = -6/(2 - x) + 3, -3 = -6(2 - x), 2 - x = (-6)/(-3), 2 - x = 2, x = 0
X-intercept is 0.

To find the intervals of increase and decrease, solve for f'(x) and then set it to 0.

f(x) = -6/(2 - x) + 3
f(x) = -6 (2 - x)^(-1) + 3
f'(x) = 6 (2 -x)^(-2)
f'(x) = -6/(2 - x)^2

Setting f'(x) = 0, we get

0 = -6/(2 - x)^2

Note that critical values are those which make f(x) = 0 OR f(x) undefined. In this case, x = 2 would make the function undefined.

To find the intervals of increase and decrease, all we have to do is test one value GREATER than 2, and one value LESS than 2, for positivity/negativity.
Test 3: -6/(2 - 3)^2 = [negative].
Test 0: -6/(2 - 0)^2 = [negative].

Therefore, f(x) is decreasing on (-infinity, 2) U (2, infinity).
f(x) is NEVER increasing.
Since we obtained no critical values where f'(x) = 0, there is no relative min or relative max.

To determine intervals of concavity, we solve for the second derivative. Recall that
f'(x) = -6/(2 - x)^2. Changing this algebraically, we get
f'(x) =(-6) (2 - x)^(-2). Finding the derivative using the power rule,

f''(x) = (12) (2 - x)^(-3) [-1]
f''(x) = -12 (2 - x)^(-3)
f''(x) = -12 / (2 - x)^3

We again get x = 2 as our critical number for the second derivative. Again, we test a value greater than and a value less than 2.

Test 0: -12/(2 - 0)^3 = -12/2 = [negative]; therefore concave down on (-infinity, 2)
Test 3: -12/(2 - 3)^3 = [positive]; therefore, concave up on
(2, infinity).

f(x) is concave up on (2, infinity)
f(x) is concave down on (-infinity, 2)

There are no points of inflection (since we got no defined values for f''(x) = 0).

Answer 2

The domain of a function is the set of x values that can be plugged into the function and get a "reasonable answer". This might seem like all values would work, but in this case, if x = 2, then 2-x = 0 and with a 2-x in the denominator, you get an invalid (infinite) result.
So, the domain of this function is every number except 2.

For the range, this is the set of numbers that are the possible OUTCOMES of the equation.
Well, if x is just a LITTLE larger than 2, than 2-x is a very small positive number and f(x) is a very LARGE negative number. If x is just a little smaller than 2, then 2-x is a very small negative number and f(x) is a very large positive number. You can convince yourself of this by trying some values or graphing the equation.

Asymptotes are lines (horizontal, vertical, or slant) that the function approaches. Since things go haywire when x is near 2, that makes a vertical axymptote at x = 2 likely.

The x intercept is anywhere f(x) = 0, so set f(x) to zero and sove for x

The y intercept is simply f(0), or 0 (since -3x = 0), so this means you have found at least one x intercept!

I don't know what f(x) decreases on... or increases on... mean

A relative maximum or minimum occurs when f'(x) is zero and when you increase or decrease x by a small amount, the function increases for both conditions or decreases for both conditions. At these points, f'(x) = 0 but f''(0) does not equal zero.
At inflection points, f'(x) = 0 and f''(x) = 0

at relative maxima, f'(x) = 0 and f'''(x) is negative
at relative minima, f'(x) = 0 and f''(x) is positive

Answer 3

horizontal asymptote
vertical asymptote
slant asymptote


http://www.purplemath.com/modules/asymtote3.htm