graphing rational expressions

[maeveoneill]

for the equations (x^3 +16) /x could someone please help me with finding the critical points. (i think they are (2,12) and (0,0) and points of inflection, regions of increase decrease, and points of inflation... i have the graphh.. and my work is not going accordingly.. thank you


ahhhhhhhhh

 

[ChaoticLlama]

Are you using quotient rule? Well.. in this case, you can use power rule. Since the denominator is a monomial, you can divide both terms by x individually.

f(x) = (x³ + 16) /x
f(x) = x² + 16/x

f'(x) = 2x - 16/x²

f''(x) = 2 + 32/x³

To find the critical numbers of f(x), set f'(x) to zero and find when the derivative is zero or undefined.

To find the inflection points of f(x), set f''(x) to zero and find when the second derivative is zero or undefined.

Can you go from here?

 

[soroban]

Hello, maeveoneill!

For the function $\displaystyle f(x)\:=\:\frac{x^3\,+\,16}{x}$ find the critical points
. . . and points of inflection, intervals of increase/decrease

\(\displaystyle \text{We have: }\,f(x)\:=\:\L\frac{x^3\,+\,16}{x}\)$\displaystyle \:=\:x^2\,+\,16x^{^{-1}}$

$\displaystyle \text{Then: }\,f'(x)\:=\:2x\,-\,16x^{-2}$ . . . $\displaystyle \text{and: }\,f''(x)\:=\:2\,+\,32x^{^{-2}}$


Critical points: $\displaystyle \,f'(x)\,=\,0\;\;\Rightarrow\;\;2x\,-\,\frac{16}{x^2}\:=\:0\;\;\Rightarrow\;\;2x^3\,=\,16\;\;\Rightarrow\;\;x\,=\,2$

There is <u>one</u> critical point: $\displaystyle \,(2,\,12)$

There is a discontinuity at $\displaystyle x\,=\,0$ which must be investigated.

$\displaystyle \;\;$On $\displaystyle (-\infty,\,0):\;f'(x)$ is negative: $\displaystyle \searrow$

$\displaystyle \;\;$On $\displaystyle (0,\,2):\;\;f'(x)$ is negative: $\displaystyle \searrow$

$\displaystyle \;\;$On $\displaystyle (2,\,\infty):\;f'(x)$ is positive: $\displaystyle \nearrow$

$\displaystyle f(x)$ is decreasing on: $\displaystyle (-\infty,\,0)\,\cup\,(0,\,2)$
$\displaystyle f(x)$ is increasing on: $\displaystyle (2,\,\infty)$


Inflection points: $\displaystyle \,f"(x)\,=\,0\;\;\Rightarrow\;\;2\,+\,\frac{32}{x^3}\:=\:0\;\;\Rightarrow\;\;x\,=\,-\sqrt[3]{16}$

There is one inflection point: $\displaystyle \,(-\sqrt[3]{16},\,0)$

On $\displaystyle (-\infty,\,-\sqrt[3]{16}):\;f"(x)$ is positive: $\displaystyle \cup$

On $\displaystyle (-\sqrt[3]{16},\,0):\;f"(x)$ is negative: $\displaystyle \cap$

On $\displaystyle (0,\,\infty):\;f"(x)$ is positive: $\displaystyle \cup$[/tex]


I think the graph looks like this:

                |*           *
                |          *
                + *      *
    *           +    **
                +
     *          +
    ----*-------+---------------
            *   +
              * +
                |
               *|
                |