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...

f(x) = 9xtan(x)       −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.

Expert Answers
embizze eNotes educator| Certified Educator

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)