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

Question

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

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).

I posted this question earlier, however, nobody was able to come up with the correct answer. Good luck to all and thank you for your help.

Answer 1

f(x) = x^2 / (4 - x^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(4 - x^2) - x^2 (-2x)] / (4 - x^2)^2
f'(x) = [8x - 2x^3 + 2x^3] / (4 - x^2)^2
f'(x) = 8x/(4 - x^2)^2

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

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

Equating the numerator to 0 is where f'(x) = 0.
8x = 0, implies x = 0

Equating the denominator to 0 is where f'(x) is undefined.
4 - x^2 = 0 implies x = {-2, 2}

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

. . . . . . . . . (-2). . . . . . . . . . . .(0) . . . . . . . . . . . (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 = -3. Then, since
f'(x) = 8x/(4 - x^2)^2
f'(-2) = [negative]/[positive] = [negative]. Mark the region as negative.

. . . . {-} . . . (-2). . . . . . . . . . . .(0) . . . . . . . . . . . (2) . . . . . . . . .

For the second region, test x = (-1). Then
f'(-1) = [negative]/[positive] = negative. Mark the region as negative.

. . . . {-} . . . (-2). . . . . .{-}. . . . .(0) . . . . . . . . . . . (2) . . . . . . . . .

For the third region, test x = 1.
f'(1) = [positive]/[positive] = positive.
Mark the region as positive.

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


For the fourth region, test x = 3.
f'(3) = [positive]/[positive] = positive.
Mark the region as positive.

. . . . {-} . . . (-2). . . . . .{-}. . . . .(0) . . . . . {+}. . . . . (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 [0, 2) U (2, infinity)
f(x) is decreasing from (-infinity, -2) U (-2, 0]

Note that 2 and -2 have no square bracket because those values are 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 decreases up to that point, and then increases, making it a local min.

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

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

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

4 - x^2 = 0, (2 - x)(2 + x) = 0, so x = {-2, 2}

Therefore, we have a vertical asymptote at x = -2 and x = 2.

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

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

Divide top and bottom by x^2,

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

Plugging in the limit directly, we get
[1 / [0 - 1]) = 1/(-1) = -1.

Therefore, we have a horizontal asymptote at y = -1.

d) To find intervals of concavity, find the second derivative and repeat the process.

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

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

f''(x) = [8(16 - 8x^2 + x^4) - 8x( -8x + 2x^3)] / [4 - x^2]^4

f''(x) = 8[(16 - 8x^2 + x^4) - x(-8x + 2x^3)] / [4 - x^2]^4

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

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

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

Equate f''(x) to 0,

0 = 8[x^4 - 2x^4 + 16] / [4 - x^2]^4
0 = 8[16 - x^4] / [4 - x^2]^4
0 = 8[ (4 - x^2)(4 + x^2) ] / [ (4 - x^2)^4
0 = 8[ (4 + x^2) / (4 - x^4)^3 ]

Repeat the process for intervals of increase/decrease.
There will be no solution for equating the numerator to 0, but for the denominator,
[4 - x^2]^4 = 0, 4 - x^2 = 0, x = 2, -2.

Make your number line.

. . . . . . . . (-2) . . . . . . . . . . (2) . . . . . . . . . .

For f''(x) = 8[ (4 + x^2) / (4 - x^4)^3 ]

f''(-3) = [positive] / (4 - 81) = positive/negative = negative.

. . . .{-} . . . (-2) . . . . . . . . . . (2) . . . . . . . . . .

f''(0) = [positive]/[positive] = positive.

. . . .{-} . . . (-2) . . . . {+}. . . . . (2) . . . . . . . . . .

f''(3) = [positive]/(4 - 81) = positive/negative = negative.

. . . .{-} . . . (-2) . . . . {+}. . . . . (2) . . . {-} . . . . .

The positive regions are intervals of concave up.
The negative regions are intervals of concave down.

Concave up on (-2, 2)
Concave down on (-infinity, -2) U (2, infinity)

There are no inflection points because the second derivative has no solutions for where f''(x) = 0.

*** By the way, the question you posted earlier was posted incorrectly, hence not getting the correct answer ***