Homework # 2

**^_^Engineer_Adam^_^** · February 7th 2007, 03:20 AM

Plot the graph of the function and from the graph, estimate the point of inflection and where the graph is concave upward and concave downward.
Confirm your estimates analytically.
$f(x) = (x + 2)^{\frac{1}{3}}$

**Soroban** · February 7th 2007, 06:43 AM

Hello, ^_^Engineer_Adam^_^!

Plot the graph of the function and from the graph, estimate the point of inflection
and where the graph is concave upward and concave downward.
Confirm your estimates analytically.
. . $f(x) = (x + 2)^{\frac{1}{3}}$

I used an "eyeball" approach to graph it.

We have: . $y \:=\:(x + 2)^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:y^3 - 2$

This is a "sideways" cubic with its vertex at (-2,0).

Code:

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

It appears that it is concave up for $x < -2$ and concave down for $x > -2$ .

We have: . $f'(x) \:=\:\frac{1}{3}(x + 2)^{-\frac{2}{3}}$

Then: . $f''(x)\:=\:-\frac{4}{9}(x + 2)^{-\frac{5}{3}} \:=\:-\frac{4}{9(x+2)^{\frac{5}{3}}}$

At $x = \text{-}2\!:\:f''(-2)$ is undefined.

For $x < \text{-}2\!:\;f''(\text{-}3) \:=\:-\frac{4}{9(\text{-}1)^{\frac{5}{3}}} \:=\:+\frac{4}{9}$ . concave up: $\cup$

For $x > \text{-}2\!:\;f''(\text{-}1) \:=\:-\frac{4}{9(1)^{\frac{5}{3}}} \:=\:-\frac{4}{9}$ . concave down: $\cap$

Therefore: . $\begin{array}{cc}\text{concave up} & \text{on }(\text{-}\infty,\,\text{-}2) \\ \text{concave down} & \text{on }(\text{-}2,\,\infty) \end{array}$

**AfterShock** · February 10th 2007, 07:11 PM

Originally Posted by Soroban

Hello, ^_^Engineer_Adam^_^!

I used an "eyeball" approach to graph it.

We have: . $y \:=\<img src=$ x + 2)^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:y^3 - 2" alt="y \:=\

x + 2)^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:y^3 - 2" />

This is a "sideways" cubic with its vertex at (-2,0).

Code:

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

It appears that it is concave up for $x < -2$ and concave down for $x > -2$ .

We have: . $f'(x) \:=\:\frac{1}{3}(x + 2)^{-\frac{2}{3}}$

Then: . $f''(x)\:=\:-\frac{4}{9}(x + 2)^{-\frac{5}{3}} \:=\:-\frac{4}{9(x+2)^{\frac{5}{3}}}$

At $x = \text{-}2\!:\:f''(-2)$ is undefined.

For $x < \text{-}2\!:\;f''(\text{-}3) \:=\:-\frac{4}{9(\text{-}1)^{\frac{5}{3}}} \:=\:+\frac{4}{9}$ . concave up: $\cup$

For $x > \text{-}2\!:\;f''(\text{-}1) \:=\:-\frac{4}{9(1)^{\frac{5}{3}}} \:=\:-\frac{4}{9}$ . concave down: $\cap$

Therefore: . $\begin{array}{cc}\text{concave up} & \text{on }(\text{-}\infty,\,\text{-}2) \\ \text{concave down} & \text{on }(\text{-}2,\,\infty) \end{array}$

Your second derivative should be:

-2/(9(x + 2)^5/3)

But, it did not change the results;

f''(-1) = -2/9
f''(-3) = 2/9,
f''(-2) still undef

Thus, we have a change in concavity at x = -2

And CD from -2 to inf; CU from -inf to -2.

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