# Math Help - monotonicity and concavity

1. ## monotonicity and concavity

so the problem is to determine where each function is increasing, decreasing, concave up, and concave down for this equation:
y=1/(1+e^-x), x=all real numbers
for this one, honestly, i have no idea how to even start.

second problem is y=(2x-3)^(1/3)
i took out the first derivative and got 2/3(2x-3)^(-2/3)
but i don't know how to get the second derivative right.

thanks!

2. Hello, katieeej!

You're expected to know about the derivatives:

. . $\begin{array}{ccc} y' > 0 & \text{increasing} & ^{\nearrow}\\ y' < 0 & \text{decreasing} & _{\searrow}\\ \\ y'' > 0 & \text{concave up} & \cup \\ y'' < 0 & \text{concave down} & \cap \end{array}$

Determine where each function is increasing, decreasing, concave up, and concave down.

$(1)\;\;y\:=\:\frac{1}{1+e^{-x}}$
This one is easier to differentiate if we eliminate the negative exponent.

Multiply top and bottom by $e^x\!:\quad y \:=\:\frac{e^x}{e^x+1}$

First derivative: . $y' \:=\:\frac{(e^x+1)e^x - e^x(e^x)}{(e^x+1)^2} \:=\:\frac{e^x}{(e^x+1)^2}$

This expression is always positive.
. . The graph is always increasing.

. . $\boxed{\begin{array}{cc}\text{Increasing:}&(-\infty, \infty) \\\text{Decreasing:}&\text{Never}\end{array}}$

Second derivative: . $y'' \:=\:\frac{(e^x+1)^2\!\cdot\!e^x - e^x\!\cdot\!2(e^x+1)\!\cdot\!e^x}{(e^x+1)^4} \:=\:\frac{1-e^x}{(e^x+1)^3}$

Inflection point when $y'' \,=\,0\!:\;\;1-e^x \:=\:0 \quad\Rightarrow\quad e^x \;+\:1 \quad\Rightarrow\quad x \:=\:0$

If $x = -1\!: \;\;y'' \:=\:\frac{1-\frac{1}{e}}{(\frac{1}{e} + 1)^3} \:=\:\frac{(+)}{(+)} \:=\:(+)$ . . . positive, concave up

If $x = 1\!:\;\;y'' \:=\:\frac{1-e}{(e+1)^3} \:=\:\frac{(-)}{(+)} \:=\:(-)$ . . . negative, concave down

. . $\boxed{ \begin{array}{cc}\text{concave up:} & (-\infty,0) \\ \text{concave down:} & (0,\infty) \end{array}}$