Continuity calculus problems?

Question

Question: Show that "f" is continuous on (-oo, oo)

f(x) = { sinx if x < pi/4 , cosx if x > or = pi/4 }

I'm really stuck. I don't know how I should work this problem.

Any help is appreciated. Thanks.

Answer 1

at x = pi/4, sin x = cos x = 1/sqrt(2)

since both sinx and cos x are continuous functions, and they match at x = pi/4, the defined function is continuously everywhere.

The only thing to watch out is the derivative of the defined function is disccontinuos at x =pi/4

Answer 2

We see that on (-oo,pi/4), sin(x) is continuous, as well as cos(x) on (pi/4,oo). So, the only point at which continuity is an issue is pi/4. We typically do these by showing the left and right limits of f(x) are equal and equal to f(pi/4), meaning does sin(x) = cos(x) at pi/4. If it does, then the function's curve remains connected and (you guessed it, continuous). Well cos(pi/4) = 1/sqrt(2) and sin(pi/4)=1/sqrt(2). Yay! It's continuous. If you need to use limits, then show that lim sin(x) = 1\sqrt(2) as x approaches pi/4 from the left and lim cos(x) = 1\sqrt(2) as x approachs pi/4 from the right.

Answer 3

First both sin and co are continuous over their respective domains, x< pi/4 and x > =pi/4 respectively. You have to show that at pi/4 both sin(x) and cos(x) give the same value - they do sqrt(2)/2.

Answer 4

If I read this correctly, you are trying to prove the continuity of f ( x ), not f ' ( x ).

sine and cosine are continuous for all real values x, but not so sure at pi / 4, since this is where they are to be "glued together", and did they do it correctly ???? ........ we will have to check continuity there......

need to check 3 things..........
1. lim x --> pi / 4 must exist
2. f ( pi / 4) must exist
3. ans 2 must = ans to 1

1. left hand limit.....Lim x---> ( pi / 4) - [ sin(x) ] = sin( pi / 4 ) = sqrt( 2 ) / 2
RHL ...Lim x ---> (pi / 4 ) + [ cos(x) ] = cos(pi/4) = sqrt( 2 ) / 2....so limit as x ---> pi / 4 is sqrt ( 2 ) / 2

2. f (pi/4) = cos(pi / 4 ) = sqrt ( 2 ) / 2

3. notice both answers to 1 , 2 are the same.....thus f ( x) is continuous at pi / 4

Answer 5

I enjoyed those in Calculus. those are actual notably basic, yet you may desire to understand a thank you to decode them. first, you may desire to boost the equation by way of fact the discontinuity creates a nil/0 answer. All you may desire to do is get the two equations (Num. and Denom.) into their root varieties. for the numerator, it would of direction be (x+a million) (x-a million) the denominator is basically (x+a million). then cancel out any like words, thus it is the (x+a million) once you cancel that out you wind up with basically (X-a million) which supplies -2 this is the single y-fee that truly does not exist. and additionally with the aid of watching the equation, you may understand that it will notably much be a on the instant line, different than that the element (-a million,-2) does not exist by way of fact we are clever human beings and basically figured that out with the above. this a hollow as antagonistic to a vertical asymptote/endless discontinuity. And on the subject of the above assertion, no, this isn't a on the instant line and should no longer be called so. this is a discontinuous Rational equation, yet from a primary look it does seem to be a on the instant line. this is why in calculus you may desire to by no potential count fullyyt on a graph of an equation to be certain issues. keep in mind: all factors are infinitesimally products which at the instant are and returned undetectable with the bare eye.

Answer 6

Focus at the point where it appears to be discontinuous, i.e., pi/4. Use either geometry or existence of derivative at the point to show continuity. At all other places, it is trivial to show continuity.