Question #e40e5

1 Answer
Nov 16, 2017

The function is convex on (-pi/2,0)uu(pi/2,pi)(π2,0)(π2,π) and is concave on (-pi,-pi/2)uu(0,pi/2)(π,π2)(0,π2).

Explanation:

First, note that ff is convex at x=ax=a when f''(a)>0 and f is concave at x=a when f''(a)<0.

So, we need to find this function's second derivative and then examine when it's positive (the function is convex) and when the second derivative is negative (the function is concave).

We'll use the trigonometric derivatives d/dxtanx=sec^2x and d/dxsecx=secxtanx.

f(x)=2x-tanx

f'(x)=2-(secx)^2

To find the next derivative, we'll need to use the chain rule.

f''(x)=-2(secx)^(2-1)d/dxsecx=-2secx(secxtanx)=-2sec^2xtanx

It will be easier to think about when this is positive and negative to think about it in basic terms: sine and cosine.

f''(x)=-2 1/cos^2xsinx/cosx=(-2sinx)/cos^3x

Let's examine each quadrant:

  • "QIII"

Here, sinx<0 and cosx<0. (Note then that cos^3x<0 as well.)

Using a very rough notation in which I only denote positives and negatives for (-2sinx)/cos^3x, we see that

f''(x)=(-(-))/(-)=-

So f is concave on -pi lt x lt -pi/2.

  • "QIV"

Here, sinx<0 and cosx>0, so cos^3x>0. Then

f''(x)=(-(-))/(+)=+

So f is convex on -pi/2 lt x lt 0.

  • "QI"

Wherein sinxgt0,cosxgt0,cos^3x>0, so

f''(x)=(-(+))/(+)=-

And f is concave on 0ltxltpi/2.

  • "QII"

Here sinx>0 but cosx<0,cos^3x<0, so

f''(x)=(-(-))/(+)=+

So f is convex on pi/2 lt x lt pi.