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Let f be the function defined by f(x)= (x+sinx)/(cosx) for -pi/2<x<pi/2state...

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hellome234 | (Level 1) Salutatorian

Posted February 14, 2012 at 6:34 PM via web

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Let f be the function defined by f(x)= (x+sinx)/(cosx) for -pi/2<x<pi/2

state whether f is an even or odd function. Determine f'(x) write an equation of the line tangent to the graph of f at the point where x=0

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nathanshields | High School Teacher | (Level 1) Associate Educator

Posted February 14, 2012 at 8:24 PM (Answer #1)

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Remember what even and odd functions mean.  Even: f(-x)=f(x). Odd: f(-x)=-f(x). Let's check it out:

`f(-x)=(-x+sin(-x))/(cos(-x))`

Now, sin(-x) = -sin(x), but cos(-x)=cos(x), so we have

`=(-x-sin(x))/cos(x)=-(x+sin(x))/cos(x)` , which is -f(x).

This is the definition of an odd function.

Derivative is going to require quotient rule...

(d(top)*bottom - d(bottom)*top)/(bottom squared)

`((1+cos(x))*cos(x)-(-sin(x))*(x+sinx))/(cos^2(x))`

Simplifying a little we have

`(cos+cos^2+xsin+sin^2)/cos^2`

remember that `cos^2+sin^2=1` (I know I'm being sloppy with the notation, but I'm trying to keep it simple).

=`(1+cos(x)+xsin(x))/(cos^2(x))`

Evaluating at x = 0, we have

`(1+1+0)/(1)=2`

So this is the slope of the tangent line when x=0.  To find the equation, we need a point on the function.  f(0)=0/1=0, so we can use the point (0,0):

Using point-slope form is always the quickest:

y-0=2(x-0) or y=2x

 

 

 

 

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