# given` f(x)=6-tan(x/2)` over the interval `(-pi,pi)` use sign analysis of the first and second derivatives to determine (in interval notation) where1. f(x) is increasing2. f(x) is decreasing3. f(x)...

given` f(x)=6-tan(x/2)` over the interval `(-pi,pi)`

use sign analysis of the first and second derivatives to determine (in interval notation) where

1. f(x) is increasing

2. f(x) is decreasing

3. f(x) is concave up

4. f(x) is concave down

5. inflection point of x= ___

*print*Print*list*Cite

The first derivative of the given function:

`f'(x) = (6-tan(x/2))' = -1/2 sec^2 (x/2)`

**This is expression is not positive for any value of x, so the function is decreasing everywhere on `(-pi, pi)` .**

The second derivative of the given function is

`f''(x) = (-1/2 sec^2(x/2))' = -1/2 * 2sec(x/2)*1/2 tan(x/2) sec(x/2)=`

`=-1/2 sec^2(x/2) tan(x/2)`

This expression is positive when `tan(x/2)` is negative, which is on `(-pi,0)`

``(As can be seen on the graph of `tan(x/2)` below. The values of `pi` and `-pi` are approximately 3 and -3, respectively.)

**So the the function is concave up on the interval `(-pi, 0)` .**

Similarly, the expression for f''(x) is negative when `tan(x/2)` is positive, which is on `(0, pi)` , as can be seen on the graph above.

**So the the function is concave down on the interval `(0, pi)` .**

**The inflection point of the graph is where f''(x) is 0, which is at x = 0.**