# f(x) = 9xtan(x) −π/2 < x < π/2a. Find the interval where the function is increasing and decreasing. b. Find the local minimum value c. Find the interval where the function is...

*π*/2 < x <

*π*/2

a. Find the interval where the function is increasing and decreasing.

b. Find the local minimum value

c. Find the interval where the function is concave up.

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Given `f(x)=9xtanx` for `-pi/2<x<pi/2` :

(1) `f'(x)=9tanx+9xsec^2x`

(2) `f''(x)=9sec^2x+9sec^2x+9x(2)(secx)(secxtanx)`

`=18sec^2x+18xsec^2xtanx`

`=18sec^2x(1+xtanx)`

** In (1) we use the product rule `d/(dx)[fg(x)]=f'g+fg'`

** In (2) we use the product rule and the chain rule

(3) In order to find where the function is increasing or decreasing, we look at the sign of the first derivative. A positive first derivative indicates the function is increasing, while a negative first derivative indicates the function is decreasing.

Here `f'(x)=9tanx+9xsec^2x`

(a) `9tanx` is negative on `(-pi/2,0)` and positive on `(0,pi/2)`

(b) `9xsec^2x<0` on `(-pi/2,0)` and `9xsec^2x>0` on `(0,pi/2)`

(c) Since the sum of two real numbers maintains the sign, the first derivative is negative on `(-pi/2,0)` and positive on `(0,pi/2)`

**Thus the function decreases on `(-pi/2,0)` and increases on `(0,pi/2)` **

(4) Local extrema can only occur at critical points; i.e. where the first derivative is zero or fails to exist.

`9tanx+9xsec^2x=0 => x=0` on the indicated domain.

**Using the first derivative test we find that (0,0) is a local minimum.**

(5) To find concavity we look at the sign of the second derivative. A positive 2nd derivative indicates the function is concave up, while negative is concave down.

`f''(x)=18sec^2x(1+xtanx)`

`18sec^2x>0 forall x` ; so we are only concerned with `1+xtanx` . Now `xtanx>0` for `-pi/2<x<0` since both factors are negative, and `xtanx>0` for `0<x<pi/2` since both factors are positive, and `xtanx=0` for x=0.

Thus `18sec^2x(1+xtanx)>= 0 forall x in (-pi/2,pi/2)` .

**The function is concave up on the entire interval.**

(6)

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