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.
Determine whether the curve has a tangent at the indicated point.
f(x)= sin x 0<x<3pi/4
cos x 3pi/4<x<2pi
at x = 3pi/4
3 answers
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.
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.