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October 26, 2014

October 26, 2014

Posted by **Hannah** on Tuesday, September 7, 2010 at 11:28pm.

f(x)= sin x 0<x<3pi/4

cos x 3pi/4<x<2pi

at x = 3pi/4

- Calculus -
**bobpursley**, Wednesday, September 8, 2010 at 12:29amThe 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.

- Calculus -
**Hannah**, Wednesday, September 8, 2010 at 7:42amThank you. Can you tell me how to actually work the problem, too? Thanks again!

- Calculus -
**MathMate**, Wednesday, September 8, 2010 at 9:22amAnother 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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