# 資訊系

3. Let $f(x) = -x^{2/3}(3-x)^{1/3}$.
( a )Find $f'(x)$.
( b )Find $f''(x)$.
( c )Find all local maxima and local minima of function f if they exist.
( d )Find all inflection points of f if they exist.

**sol:**
( a )
Let $f(x) = g(x)h(x)$ and $g(x) = -x^{2/3}$, $h(x) = (3-x)^{1/3}$.
Since $f'(x) = g'(x)h(x) + g(x)h'(x)$ **(1 pts)** and $g'(x) = -\frac{2}{3}x^{-1/3}$, $h'(x) = -\frac{1}{3}(3-x)^{-2/3}$.
Hence $f'(x) = -\frac{2}{3}x^{-1/3}(3-x)^{1/3} + \frac{1}{3}x^{2/3}(3-x)^{-2/3 }$ **(1 pts)** $= \frac{x-2}{x^{\frac{1}{3}}(3-x)^{\frac{2}{3}}}$.**(1 pts)**

( b )
Let $f'(x) = \frac{k(x)}{l(x)} = \frac{x-2}{x^{\frac{1}{3}}(3-x)^{\frac{2}{3}}}$.
Since $f''(x) = \frac{k'(x)l(x) - k(x)l'(x)}{(l(x))^2}$, **(1 pts)** and $k'(x) = 1$, $l'(x) = \frac{2-x}{x^{\frac{1}{3}}(3-x)^{\frac{2}{3}}}$
then $f''(x) = \frac{(1)x^{\frac{1}{3}}(3-x)^{\frac{2}{3}} - (x-2)\frac{2-x}{x^{\frac{1}{3}}(3-x)^{\frac{2}{3}}}}{(x^{\frac{1}{3}}(3-x)^{\frac{2}{3}})^2}$ **(1 pts)** $= \frac{2}{x^{\frac{4}{3}}(3-x)^{\frac{5}{3}}}$**(1 pts)**

( c )
To find the critical numbers we set $f'(x) = 0$ and obtain $x = 0$, $x = 2$, $x = 3$. **(2 pts)**
So we have
$f'(x) > 0$ when $x \in (-\infty, 0)$.
$f'(x) < 0$ when $x \in (0, 2)$.
$f'(x) > 0$ when $x \in (2, 3)$.
$f'(x) > 0$ when $x \in (3, \infty)$.
**(4 pts)**
Because $f'(x)$ changes from positive to negative at $0$, the First Derivative Test tell us that thereis a local maximum at $0$ and the local maximum value is
$$f(0) = 0. \qquad (1 \ pts)$$
Likewise, $f'(x)$ changes from negative to positive at $2$ and so
$$f(2) = -2^{2/3} \qquad (1 \ pts)$$ is a local minimum value.

( d )
Since
$f''(x) > 0$ when $x \in (-\infty, 0)$.
$f''(x) > 0$ when $x \in (0, 3)$.
$f''(x) < 0$ when $x \in (3, \infty)$.
**(3 pts)**
So the graph of $f$ is concave upward on $(-\infty, 3)$ **(1 pts)** and the graph of $f$ is concave downward on $(3, \infty)$ by the Concavity Test. **(1 pts)**
The point $(3, 0)$ is an inflection point since the curve change from concave upward to concave downward there. **(1 pts)**

4.Let $f(x) = 2x^{5/3} - 5x^{2/3}$.
( a )Find the intervals of increase.
( b )Find the intervals of decrease.
( c )Find the intervals on which f is concave downward.
( d )Find the intervals on which f is concave upnward.

**sol:**
( a ), ( b )
$$\begin{aligned}
f'(x) &= \frac{d}{dx}(2x^{5/3}) - \frac{d}{dx}(5x^{2/3}) \\
& = \frac{10}{3}x^{2/3} - \frac{10}{3}x^{-1/3} \\
& = \frac{10}{3}x^{-1/3}(x-1) \\
&(4 \ pts)
\end{aligned}$$
To find the critical numbers we set $f'(x) = 0$ and obtain $x = 0$, $x = 1$. **(2 pts)**
So we have
$f'(x) < 0$ when $x \in (-\infty, 0)$.
$f'(x) < 0$ when $x \in (0, 1)$.
$f'(x) > 0$ when $x \in (1, \infty)$.
**(3 pts)**
Thus $f$ is increasing on $(1, \infty)$ **(1 pts)** and $f$ is decreasing on $(-\infty, 1)$ by the Increasing/Decreasing Test **(1 pts)**

( c ), ( d )
$$\begin{aligned}
f''(x) &= \frac{10}{3}\frac{d}{dx}(x^{-1/3}(x-1)) \\
& = \frac{10}{3} (\frac{-1}{3}x^{-4/3}(x-1) + x^{-1/3}(1)) \\
& = \frac{10}{9}x^{-4/3}(2x+1) \\
&(4 \ pts)
\end{aligned}$$

Since
$f''(x) < 0$ when $x \in (-\infty, -\frac{1}{2})$.
$f''(x) > 0$ when $x \in (-\frac{1}{2}, 0)$.
$f''(x) > 0$ when $x \in (0, \infty)$.
**(3 pts)**
So the graph of $f$ is concave upward on $(-\frac{1}{2}, \infty)$ **(1 pts)** and the graph of $f$ is concave downward on $(-\infty, -\frac{1}{2}).$ by the Concavity Test **(1 pts)**