If f(x)=[sinx+cosx], then the number of points of discontinuity of the function for x belongs to (0,2[math]\pi[/math]). Here brackets mean the greatest integer function.
We know, [p] is discontinuous at integral values of p.Now, sinx+cosx = √2sin(x+π/4)So, [sinx+cosx] = [√2sin(x+π/4)]-will be discontinuous at integral values of √2sin(x+π/4) in the interval (0,2π).Now, by inspection, we find that,at x = π/2 ,3π/4 , π, 3π/2 , 7π/4; the value of √2sin(x+π/4) is an integer. (Verify)So, there are 5 points in the interval (0,2π), for which [sinx+cosx] is discontinuous. (Ans)Hope, you'll understand..!!Thank You!
I want function which has no maximum or minimum on an closed interval !!!!!!!!!!!!!!!!?
I don't know
Is sin(tanx) a continuous function?
where tangent is infinite, sin(tanx) oscillates like in an infinite way. Look at the graph of it on Desmos. So is it continuous or nah? I think continuous, but I'm not sure.
How do I check the continuity and derivative of the function mod x+1 on a closed interval of -2,0?
Look,|x+1| is always Continuous in its domain …so does in [-2,0],Now for diffrentiablity of |x|, its indiffrentiable at x=0 ie when its graph take sharp curve . Hence in this case |x+1| is not diffrentiable atX+1=0 ie at x=-1Now checking in given domain1> its continous2> its indiffrentiable at x=-1
How is f(x)=tanx a continuous function?
It is continuous b/w- π/2 to π/2.Not on entire real line.
Is tan(x) a continous function?
What do we mean by continuity? No definition, just think. Is it always moving and connected, that is how I think of it. Think of a wire, we can break it if we want, otherwise it stays intact….continuous, isn’t it?Plug in [math]x=\dfrac{\pi}{2}[/math] for [math]\tan x[/math] on the calculator, do you see Mathematical Error message pop up?Okay, let’s look at the definition now.Continuity: A function [math]f(x)[/math] is said to be continuous at an arbitrary point [math]x=c[/math], if[math]\displaystyle \lim_{x\to c} f(x)[/math] exists, and[math]\displaystyle \lim_{x\to c} f(x)=f(c)[/math]Okay, we can say [math]\tan \dfrac{\pi}{2}=\text{undef.}[/math] , That is fine, but does the limit exist?Try taking the left and right limits at [math]x=\dfrac{\pi}{2}[/math][math]\displaystyle \lim_{x\to \frac{\pi}{2}^-} \tan x=\tan x=+ve[/math] value, angle is in the first quadrant.[math]\displaystyle \lim_{x\to \frac{\pi}{2}^+} \tan x=\tan x=-ve[/math] value, angle is in the second quadrant.Finally, here’s a visualSee the shape of the curve hate each other at [math]x=\dfrac{\pi}{2}[/math], that is why they do not connect. There are also other discontinuities, [math]\pi[/math] degrees before and after ANY discontinuity [math]\implies -\dfrac{\pi}{2},\dfrac{\pi}{2},\dfrac{3\pi}{2}…[/math] are all points of discontinuity.So, what is our answer? [math]\tan x[/math] is not continuous on [math]\R[/math], rather it is piecewise continuous.Thanks for the A2A
Let f(x)=min {tan x, cot x} where x belongs to the set of real numbers. What is the range, fundamental period, points of discontinuity and points of non-differentiability of f(x)?
While you draw the graph for min{tanx,cotx}, fundamental period can be found by looking for minimum interval after which graph is repeating.Points of discontinuity where graph breaks or reaches infinity.Points of non -differentiability where you will be able to draw two tangents.Typically points where you see sharp edges. You will find the common point or (solution of tanx=cotx)of graph of tanx and cotx will come under this.You can post another question ,if you want to know how to draw the graph of min{tanx,cotx}
If the function f(x) =(x+1) ^cotx is continuous x =0 , then f(0) should be defined as?
Put (x+1)^cot x = y. Thenln y = cot x.ln(x+1) = ln(x+1)/tan x, which is an indeterminate form of the type (0/0) as x tends to 0. Hence differentiating numerator & denominator separately gives cos^2 x/(x+1) which approaches 1 as x tends to 0. Hence by the rule of l'ho^pital, ln y approaches 1 as x tends to 0. Therefore limit of y as x tends to 0 is e^1 = e which should be taken as f(0) to make f continuous.
How is tanx a continuous function when it is discontinuous at pi/2 (90 degrees)?
Imagine the tangent as a tangent line to a circle centered on the XY axis.Now imagine that line rotating around the edge of the circle 2 pi radians.You see that at (1,0) and (-1,0) the tangent is vertical parallel with the Y axis, and horizontal at (0,1) and (0,-1). At (y=x) , y = -x, (4 areas) you get pi/4 (45deg) tangents.So even though[math]tan(x) = \frac{sin(x)}{cos(x)}[/math]is undefined when [math]cos(x) = 0[/math], it is continuous.It has a value for all x and a continuous limit.The tangent function goes to infinity as the cosine function goes to zero, but it describes the slope equation of a line on the unit circle.Another definition is the tangent of angle x is the ration of the opposite to the adjacent lines of a right triangle.[math]tan(\theta) = \frac{y}{x} [/math]A tangent of infinity is a vertical slope.A tangent of [math]0 [/math]is a horizontal slope.A tangent of [math]1[/math] is an angle of [math]\frac{\pi}{4}[/math]A third explanation is that if[math]tan(x) = \frac{sin(x)}{cos(x)}[/math]then you have 3 continuous functions : [math] [/math]sine, cosine, and division[math] -- [/math]applied to an angle, which is a real numbersine is continuous, so is cosine, and division, with the exception of division by zero , which an undefined operation on all real numbers,all values of [math]\theta [/math]are valid, even [math]\frac{\pi}{2},[/math] where [math]cos(\theta)= 0[/math].