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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Remember what even and odd functions mean. Even: f(-x)=f(x). Odd: f(-x)=-f(x). Let's check it out:
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)
Simplifying a little we have
remember that `cos^2+sin^2=1` (I know I'm being sloppy with the notation, but I'm trying to keep it simple).
Evaluating at x = 0, we have
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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