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

[Guest]

I just want to know if I am on the right page for this problem.

. . .f(x) = (x^3 + 3x^2) / (2*(x^2 + x - 6))

Once I simplify, I get:

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

. . .f"(x) = 32 / ((2x - 4)^3)

The x- and y-intercepts are (0,0); vertical asymptote at x = 2, no horizontal asymptote, but a slant asymptote at y = x (2x - 4/x^2), using long division); a hole on the graph at x = -3.

By using f'(x), I get critical numbers: x = 4, 0, 2--------->putting them into a table

| interval   | (-inf., 0) |   (0, 2)   |   (2, 4)   | (4, inf.)  |
| test numb. |     -1     |      1     |      3     |      5     |
| sign f'(x) |      +     |      -     |      -     |      +     |
| conclusion | increasing | decreasing | decreasing | increasing |

By using f"(x), critical numbers are x = -2, 2

| interval   | (-inf., -2) | (-2,2)  | (2, inf.) |
| test numb. |     -3      |    1    |     3     |
| sign f'(x) |      -      |    -    |     +     |
| conclusion |   concave   | concave |  concave  |
|            |    down     |  down   |     up    |

inflection point at x=2

Am I understanding this stuff correctly????

 

[ChaoticLlama]

your second derivative should be

f''(x) = 4 / (x - 2)³

check your work again, and then repost your analysis.

 

[soroban]

Hello, blowfish!

I think you're doing great! . . . a couple of errors, though.

\(\displaystyle f(x) \:= \:\frac{x^3\,+\,3x^2}{2(x^2\,+\,x\,-\,6)}\)

Your work is correct, but note that the function can be reduced:

\(\displaystyle \;\;f(x)\:=\:\frac{x^2(x\,+\,3)}{2(x\,-\,2)(x\,+\,3)} \:=\:\frac{x^2}{2(x\,-\,2)}\) . . . for \(\displaystyle x\,\neq\,-3\)

which tells us immediately about the "hole".

Once I simplify, I get: . \(\displaystyle f'(x) \;= \;\frac{2x^2\,-\,8x}{(2x\,-\,4)^2}\;\;\;\;f''(x)\;=\;\frac{32}{(2x - 4)^3}\;\;\) . . . Right!

Tip: leave it in factored form.

\(\displaystyle f'(x)\:=\:\frac{2x(x - 4)}{[2(x\,-\,2)]^2}\:=\:\frac{2x(x\,-\,4)}{4(x\,-\,2)^2}\:=\:\frac{x^2\,-\,4x}{2(x\,-\,2)^2}\)

\(\displaystyle f''(x)\:=\:\frac{32}{[2(x\,-\,2)]^3}\:=\;\frac{32}{8(x\,-\,2)^3}\:=\:\frac{4}{(x\,-\,2)^3}\)

The x- and y-intercepts are (0,0); a hole on the graph at \(\displaystyle x = -3\)

Vertical asymptote at x = 2, no horizontal asymptote,
but a slant asymptote at y = x (2x - 4/x^2) ? using long division.

I get: \(\displaystyle \,y \;= \;\frac{1}{2}x\,+\,1\,+\,\frac{4x\,+\,12}{2x^2\,+\,2x\,-\,12}\)

The slant asymptote is: \(\displaystyle \,y\:=\:\frac{1}{2}x\,+\,1\)

By using \(\displaystyle f'(x)\), I get critical numbers: \(\displaystyle x\,=\,0,\;2,\;4\)
Putting them into a table:

| interval   | (-inf., 0) |   (0, 2)   |   (2, 4)   | (4, inf.)  |
| test numb. |     -1     |      1     |      3     |      5     |
| sign f'(x) |      +     |      -     |      -     |      +     |
| conclusion | increasing | decreasing | decreasing | increasing |

This is absolutely correct!

Another tip: use the factored form of the derivative.
Since we're concerned with the sign only, we can work faster.

\(\displaystyle f'(x)\:=\:\frac{x(x\,-\,4)}{2(x\,-\,2)^2}\)

At \(\displaystyle x\,=\,-1\), we have: \(\displaystyle \,f'(-1)\:=\:\frac{(-1)(-5)}{2(-3)^2}\:=\:\frac{[+]}{2[+]}\:=\:[+]\;\;\nearrow\)

At \(\displaystyle x\,=\,1\), we have: \(\displaystyle \,f'(1)\:=\:\frac{1(-3)}{2(-1)^2}\:=\:\frac{[-]}{[+]}\:=\:[-]\;\;\searrow\)

By using \(\displaystyle f''(x)\), critical numbers are \(\displaystyle x \,=\,-2,\;2\;\) . . . um

| interval   | (-inf., -2) | (-2,2)  | (2, inf.) |
| test numb. |     -3      |    1    |     3     |
| sign f'(x) |      -      |    -    |     +     |
| conclusion |   concave   | concave |  concave  |
|            |    down     |  down   |     up    |

inflection point at \(\displaystyle x\,=\,2\;\) . . . no

The only critical number for the second derivative is \(\displaystyle x = 2\)

Besides, we already know that there is a vertical asymptote there, don't we?