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
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.
To determine whether the curve has a tangent at a specific point, we need to check if the derivative of the function exists at that point.
To find the derivative of the given function f(x), we need to find the derivative separately for the two intervals.
For the interval 0 < x < 3π/4, the given function is f(x) = sin(x). The derivative of sin(x) with respect to x is cos(x). Therefore, the derivative of f(x) for this interval is f'(x) = cos(x).
For the interval 3π/4 < x < 2π, the given function is f(x) = cos(x). The derivative of cos(x) with respect to x is -sin(x). Therefore, the derivative of f(x) for this interval is f'(x) = -sin(x).
Now, we need to find the value of the derivative at x = 3π/4. For this, we need to substitute x = 3π/4 into each derivative function separately.
For the interval 0 < x < 3π/4, substituting x = 3π/4 into f'(x) = cos(x), we get f'(3π/4) = cos(3π/4) = -sqrt(2)/2.
For the interval 3π/4 < x < 2π, substituting x = 3π/4 into f'(x) = -sin(x), we get f'(3π/4) = -sin(3π/4) = -sqrt(2)/2.
Since both derivatives yield the same value of -sqrt(2)/2 at x = 3π/4, we can conclude that the curve has a tangent at this point.
Therefore, the given curve has a tangent at x = 3π/4.