We have `f(x)=2sin(x)+tan(x)-3x`

`AAx in [0,pi/2]`

how can we prove that `f'(x)gt=0`

and that `3x<=2sin(x)+tan(x)`

Expert Answers

An illustration of the letter 'A' in a speech bubbles

(1) Prove that if `f(x)=2sinx+tanx-3x` , then `f'(x)>=0` on `[0,pi/2)`

`f'(x)=2cosx+sec^2x-3` .` `` ` We can show that `f'(x)>= 0` on the interval if it is increasing on the interval and `f'(0)>=0` .

To show that `f'(x)` is increasing we find `f''(x)` :

`f''(x)=-2sinx+2sec^2xtanx=2sinx(-1+1/(cos^3x))` . Now `sinx>=0` on `[0,pi/2)` , and the minimum value for `1/(cos^3x)` is 1 on the interval. Thus the minimum value of `f''(x)` is 0 on the interval. Since the second derivative is nonnegative on the interval, the first derivative is an increasing function on the interval.

Also, `f'(0)=2+1-3=0` . Therefore `f'(x)>=0 forall x in [0,pi/2)`

(2) Show that `2sinx+tanx>=3x ` on `[0,pi/2)`

The function is increasing on the interval since the first derivative is nonnegative on the interval. Also, `f(0)=0+0-0=0` . Thus `2sinx+tanx-3x>=0` and `2sinx+tanx>=3x` as required.

` `

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial