Let f(x) = x2/4x-2?

Question

a) Find the y-intercept
b) Find all critical points of f(x), all intervals where f(x) is increasing, and where f(x) is decreasing, and determine which critical points are local maximum and which are local minimum.
c) Find any vertical and horizontal asymptotes.
d) Find all intervals where f(x) is concave up, and where f(x) is concave down and find all inflection points of f(x).
e) Sketch the graph of f(x) indicating the y-intercept, the local maximum, the local minimum, the inflection points, and the asymptotes.

Having a problem finding the correct solutions. Thank you to anyone who can provide help.

Answer 1

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

a) To find the y-intercept, set x to 0.
f(0) = 0^2 / (4(0) - 2)) = 0

b) To find the critical points, find the first derivative and make it 0.

f'(x) = [2x(4x - 2) - x^2 (4)] / (4x - 2)^2
f'(x) = [8x^2 - 4x - 8x^2] / (4x - 2)^2
f'(x) = (-4x)/(4x - 2)^2

Critical points are where f'(x) = 0 or where f'(x) is undefined.
Equating f'(x) to 0,

0 = (-4x)/(4x - 2)^2

Equating the numerator to 0 is where f'(x) = 0.
-4x = 0, implies x = 0

Equating the denominator to 0 is where f'(x) is undefined.
4x - 2 = 0 implies x = 2/4 = 1/2.

Critical values: 0, 1/2. Make a number line consisting of these values.

. . . . . . . . . . . (0). . . . . . . . . . . .(1/2) . . . . . . . . . . . .

For f'(x), now we want to TEST each region for positivity; negativitity. All we need to do is test a single value.

For the first region, test x = -1. Then, since
f'(x) = (-4x)/(4x - 2)^2
f'(-1) = (-4)(-1) / (4*(-1) - 2))^2 = 4/36, which is positive. Mark the region as positive.

. . . . .{+} . . . . . (0). . . . . . . . . . . .(1/2) . . . . . . . . . . . .

For the second region, test x = (1/4). Then
f'(1/4) = (-4*1/4)/[ a positive number] = -1 divided by a positive number, which is negative. Mark the second region as negative.

. . . . .{+} . . . . . (0). . . . {-} . . . . . .(1/2) . . . . . . . . . . . .

For the third region, test x = 10000000. The beauty of this method is we can test ANY value in that region. We will get
f'(1000000) = (-4 * [something large]) / (something squared)
= (something negative)/(something positive,since something squared is always positive) = negative.
Mark the region as negative.

. . . . .{+} . . . . . (0). . . . {-} . . . . . .(1/2) . . . . . .{-} . . . . .

The regions marked as positive is where the function is increasing (be sure to include defined critical points).
The regions marked as negative is where the function is decreasing.

f(x) is increasing from (-infinity, 0]
f(x) is decreasing from [0, 1/2) U (1/2, infinity)

Note that 1/2 has no square bracket because it's what makes f'(x) undefined.

To find local min/max points, first note that our only defined point is at x = 0. At x = 0, the function increases up to that point, and then decreases, making it a local max.

f(0) = 0, so we have a local max at (0, 0).

We have no local minimum because the graph doesn't have a defined critical point that alternates from decreasing to increasing.

c) To find the vertical asymptotes, all you have to do is equate the denominator of f(x) to 0.

4x - 2 = 0, 4x = 2, x = 2/4 = 1/2

Therefore, we have a vertical asymptote at x = 1/2.

To find the horizontal asymptotes, you need to find the limit as x approaches infinity, and x approaches negative infinity.

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

Divide top and bottom by x^2,

lim [1 / [4/x - 2/x^2] ]
x -> infinity

This gives us the form [1/0], upon plugging in infinity, so the limit doesn't exist. There are no horizontal asymptotes.

d) To find concavity, obtain the second derivative. As a reminder,
f'(x) = (-4x)/(4x - 2)^2
f''(x) = [(-4)(4x - 2) - (-4x)(4)] / (4x - 2)^4

f''(x) = [-16x + 8 + 16x] / (4x - 2)^4
f''(x) = 8/(4x - 2)^4

Solve for critical points once again.

0 = 8/(4x - 2)^4

Critical points is what makes f''(x) = 0 or undefined. In this case, we have no value that makes f''(x) = 0, but we do have one that makes it undefined.
(4x - 2)^4 = 0
4x - 2 = 0
x = 1/2

Make a number line again.

. . . . . . . . . . . (1/2) . . . . . . . . . . . .

For f''(x), test x = 0.
f''(x) = 8/(4x - 2)^4
f''(0) = 8/(4(0) - 2)^4 = 8/(-2)^4 = 8/16 = 1/2, which is positive.

. . . . .{+} . . . . . (1/2) . . . . . . . . . . . .

Test x = 1.
f''(1) = 8/(4*1 - 2) = 8/(2) = 4, which is positive.

. . . . .{+} . . . . . (1/2) . . . . .{+} . . . . . .

The positive intervals are where f(x) is concave up.
The negative intervals are where f(x) is concave down.
Unlike intervals of increase/decrease, we ALWAYS exclude the critical values when finding the interval, regardless of whether it made f''(x) = 0 or undefined.

Concave up on (-infinity, 1/2) U (1/2, infinity)
There are no inflection points, because we have no defined critical points.

e) I can't do this part for you. Just start by making your graph, graphing a "dot" at (0, 0), graphed your dashed asymptote at x = 1/2, knowing the behaviour given all of the above information, and so forth.

Answer 2

Or simply remember that if y = ax^2 + bx + c, the x coordinate of the vertex is x = -b / (2a) So x = - (-4) / 2(1) = -4/2 = 2 then just plug 2 in for x to get the y coordinate. No need to make this a calculus or even an algebra II problem!!

Answer 3

f(x) = (x^2)/(4x) - 2 = x/4 - 2 a straight line
a) the y-intercept is -2
b) its a straight line, minimum -infinity, max + infinity
c) no asymptotes
d) see above
e) see above

Answer 4

We're not here to do your homework.