# 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)`

(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.**

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