Deduce f(x) is postive for x>0

Can you see that if you can show that the derivative is never negative, then you can conclude that the function is positive for x>0?

Now look for an algebraic way to show that. I think what you did with the quadratic function probably came very close.
 
Dear Dr Peterson

I can that the derivative cannot ne negative as when cos x is a maximum value , 4 cos x/2 will be a minimum, and hence the minimum value of the function as a whole is zero. I can input values of x and show that but not prove it for all values of x. I worked out the value of x as a stationary point.
Do I set the derivative as an inequality aand prove it cannot be the case?

Also proving that the derivative of a function is always positive, I am unclear how this demonstrates the function is positive for values of x >0.

Thanks for help.
 
You wind up with the quadratic:

[MATH]f'(x)=\cos^2\left(\frac{x}{2}\right)-2\cos\left(\frac{x}{2}\right)+1=0[/MATH]
Or:

[MATH]f'(x)=\left(\cos\left(\frac{x}{2}\right)-1\right)^2=0[/MATH]
This means the function is increasing everywhere except as you found for [MATH]x=4\pi k[/MATH] where k∈Zk\in\mathbb{Z}. What you found also implies the function will never be decreasing.

We know f(0)=f′(0)=0f(0)=f'(0)=0, but as soon as xx increases beyond zero, the derivative must be positive, and so the function is immediately increasing, and thereafter will never decrease, thus the function must always be positive for 0<x0<x.
 
I want to repeat one of MarkFl's point since it might have been missed with all the fine work he showed.

You need to show that f(0)>0 and that f'(x) >0 for x>0. The 2nd part, f'(x)>0, tells us that f(x) is always increasing. In the end it must be that f(x)>0 for all x>0.
 
Thank you so much for clear explanations. Demonstrate it is an increasing function (dy/dx>0), show if at one point f(0) and dy/dx are zero, then values of x beyond this mean f(x)>0 will be positive.
 
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