Posted by Hannah on Tuesday, September 7, 2010 at 11:28pm.
The real question is the f' continous at the point.
below the point: f'=cos x, at the point, f'=-.707
f' above the point; f'=-sinx, at the point, f'=-.707
so, the derivative curve is continous, and the curve has a tangent.
Thank you. Can you tell me how to actually work the problem, too? Thanks again!
Another condition for tangency at x0=(3/4)π is that f(x0) must exist.
In this particular case,
Lim f(x0-)=(√2)/2 and
Lim f(x0+)=-(√2)/2
so f(x) is discontinuous at x=x0.
In fact, f(x) is undefined at x=x0, or x0 is not in the domain of f(x).
The tangent does not exist at x0, even though the derivative (slope) is continuous throughout the function.
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