# Math Help - Homework # 2

1. ## Homework # 2

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.
$f(x) = (x + 2)^{\frac{1}{3}}$

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.
. . $f(x) = (x + 2)^{\frac{1}{3}}$
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).
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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: . $\begin{array}{cc}\text{concave up} & \text{on }(\text{-}\infty,\,\text{-}2) \\ \text{concave down} & \text{on }(\text{-}2,\,\infty) \end{array}$

3. Originally Posted by Soroban

I used an "eyeball" approach to graph it.

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

-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.