Calculus
posted by Hannah .
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.
Respond to this Question
Similar Questions

PreCalculuscheck answers
State the period and phase shift of the function y=4tan(1/2x + 3pi/8) a) 2pi, 3pi/4 b) pi, 3pi/8 c) 2pi, 3pi/8 d) pi, 3pi/8 Answer: d 2) What is the equation for the inverse of y=cos x+3: a) y=Arccos(x+3) b) y=Arccos x3 c) y=Arccos … 
Math
Test for symmetry with repsect to Q=pi/2, the polar axis, and the pole. r=16cos 3Q r=16cos(3pi/3 + 3h) = 16(cos3pi/2 3h sin 3pi/2 sin h) cos 3pi/2 = 0 sin 3pi/2 = 1 Not symmetrical My teacher said that this was wrong!Why? 
Math
Test for symmetry with repsect to Q=pi/2, the polar axis, and the pole. r=16cos 3Q r=16cos(3pi/3 + 3h) = 16(cos3pi/2 3h sin 3pi/2 sin h) cos 3pi/2 = 0 sin 3pi/2 = 1 Not symmetrical My teacher said that this was wrong!Why? 
maths
Please can you help me with this question? 
trig
Solve the equation for cos thetatan theta=0 for greater than or egual to zero but less than 2pi. Write your answer as a multiple of pi, if possible form the following choices: pi , 5pi A.   4___ 4 B. pi 3pi 5pi 7pi  ' … 
Calculus
Find the slope of the tangent line to the curve 2sin(x) + 6cos(y)  6sin(x)cos(y) + x = 3pi at the point (3pi, 7pi/2) Thank you very much for your help. 
trig
Solve cos x1 = sin^2 x Find all solutions on the interval [0,2pi) a. x=pi, x=pi/2, x= 2pi/3 b. x=3pi/7, x=pi/2, x=2pi/3 c. x=3pi/7, x=3pi/2, x=3pi/2 d. x=pi, x=pi/2, x=3pi/2 
Calculus
1) The period of a trig. function y=sin kx is 2pi/k. Then period of y=sin^2(pi.x/a) should be 2pi/(pi/a)=2a, but somewhere it is given as a. Which is correct? 
Trig
Find all solutions of the equation in the interval [0,2pi) 2 cos^2 xcos x = 0 2cos^2 + cosx + 0 (x+1/2) (x+0/2) (2x+1) (x+0) 1/2,0 2Pi/3, 4pi/3, pi/2, 3pi/2 my teacher circled pi/2 and 3pi/2 What did I do wrong? 
Algebra 2
What values for theta(0 <= theta <= 2pi) satisfy the equation?