Answered By Abdullah A

So we are given the domain (x values), which ranges from -pi/2 to pi/2.

Next we should look for asymptotes. The only thing that can result in an asymptote is the tanx. tanx =sinx/cosx. This will be discountinous when cosx=0 since that'll result in a 0 denominator. 

cosx is 0 at x=pi/2+n*pi. So this means it's discontinous at pi/2 and -pi/2. At these points we'll have vertical asymptotes. 

For increeasing/ decreasing intervals we need the derivative:

f(x)=2x-tan(x)

f'(x)=2-sec²(x)

Now we need to find the critical points, the points where the derivative is equal to 0. 

0= 2-sec²(x)

2=sec²(x)

 $\sqrt{2}$√2 =sec(x) --> sec(x) is just 1/cos(x)

 $\sqrt{2}$√2 =1/cos(x)

cos(x)=1/ $\sqrt{2}$√2  --> rationalize the denominator

cos(x)= $\frac{\sqrt{2}}{2}$√22    --> cos(x) is equal to this at pi/4 and -pi/4 (check the unit circle)

x= $\frac{pi}{4},-\frac{pi}{4}$pi4 ,−pi4  

so now we have our Critical points. We need to test around them to see the sign of the derivative to know how its changing. 

So I like to set up a number line with the critical points

 _______$-\frac{pi}{4}$‍−pi4  _________0_________ $\frac{pi}{4}$pi4  ________

Test before -pi/4, after -pi/4, and after pi/4

so lets take -pi/3, 0, and pi/3

f'(-pi/3)=2-sec²(-pi/3) --->2- [1/(cos(-pi/3))²]---> cos(-pi/3) is 0.5 (check the unit circle)

f'(-pi/3)=2-(1/0.25)

f'(-pi/3)=-2  --> derivative is negative

 

now lets test x=0

f'(0)=2-[1/(cos(0)²]--> cos(0) is 1

f'(0)=2-1

f'(0)=1   --> derivative is positive

 

now lets test x=pi/3

after plugging it in we see its negative

so at -pi/4 our derivative changes from negative to positive (which means a local min)

at pi/4 it changes from positive to negative (which means a local max) 

So now we can get the values of the max/min by plkugging in the critical points back into the original function. 

f(-pi/4)=2(-pi/4)-tan(-pi/4)

= $-\frac{pi}{2}$−‍pi2  +1

similarly 

f(pi/4)= $\frac{pi}{2}+1$pi2 +1 

 

Now we have to figure out the points of inflection. For that we need the second derivative.

so f'(x) was 2-sec²(x)

therefore f"(x)=-2tan(x)sec²(x)

to get the inflection points we need to find where the second derivative equals 0

0=-tan(x)sec²(x)

tan(x) equals 0 at x=0+pi*n. Sec(x) doesnt ever equal zero so we'll leave it alone. Our possible inflection point is x=0 since thats within our domain. 

we just need to test around x=0 to see if a sign change occurs. 

so lets try -pi/3 and pi/3

f"(-pi/3)=2*tan(-pi/3)*sec²(-pi/3)

=2*( $-\sqrt{3}$−√3 )*4

--> negative

lets test x-pi/3

f"(pi/3)=2*tan(pi/3)*sec²(pi/3)

=2*(  $\sqrt{3}$√3  )*4

--> positive

so we have a sign change from negative to positive at x=0. This means 0 is an inflection points, and that from -pi/2 to 0 we are concave down, and from 0 to pi/2 we are concave up. 

So the only thing we need now are the x and y interscepts. 

x interscept is when y=0

0=2x-tan(x)

2x=tan(x) --> the most obvious solution is x=0. apart from that you wont be able to calculate any other points without some commplex apprimations (which we dont need to do,  x=0 is more than enough for this question)

so the x-int is at x=0

and the y-int is also at y=0

now we have all the info we need to sketch the graph. 

You can see all the info we gathered above follows what the graph looks like. x,y int are at 0. Derivative is negative from -pi/2 to -pi/4, positive till pi/4, and then negatiuve again. Vertical asymptotes at -pi/2 and pi/2. Hope this helps !