f(x)=2sinx-cosx
-----------
1+tanx
In how many points in the interval [0,2pi] is the function above discontinious ?
I say 3 points book says 2.
pi/2:tangent is undefined
tan135:denominator becomes 0
tan315: denominator becomes 0.
It tends to zero from both sides...
From the left ($\pi/2$), $\tan x $ tends to $+ \infty$. From the right, $\tan x $ tends to $-\infty$.
So $$\lim_{x \rightarrow \frac{\pi}{2}^-} f(x)= \frac{2}{1+\infty}=0$$
and
$$\lim_{x \rightarrow \frac{\pi}{2}^+} f(x)= \frac{2}{1-\infty}=0$$
So from this we can say that $$\lim_{x \rightarrow \frac{\pi}{2}} f(x) =0$$.
Now a function is continuous at a point $a$ if $$\lim_{x \rightarrow a} f(x) = f(a)$$
Can you re-write $f(x)$ so that you can evaluate it at $x=\pi /2$? Hint: Use $\tan x = \frac{ \sin x}{\cos x}$ and write $1+ \tan x = \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}$ .
Kastamonu, Ahri is offering a different function that is continuous at $\frac{\pi}{2}$. The function you are given is undefined at that point, so you are correct that is it discontinuous. The definition of continuous requires the function be defined AND that the limit from the left and from the right be equal to the value defined.
There will be four points of discontinuity on the given interval.
$\dfrac{\pi}{2},\dfrac{3\pi}{4},\dfrac{3\pi}{2}, \dfrac{7\pi}{4}$
I was going on $$f\left(x\right)=\frac{2\sin x - \cos x}{1+ \tan x}=\frac{2\sin x - \cos x}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}=\frac{2\sin x - \cos x}{\frac{1}{\cos x}\left(\cos x + \sin x\right)}=\frac{\cos x \left(2\sin x - \cos x\right)}{\cos x + \sin x} $$
which can be evaluated at $x= \frac{\pi}{2}$
f(x)=(2sinx-cosx)/(1+tanx)
The function is like that.
tan135:denominator becomes 0
tan315: denominator becomes 0.
At these points it is discontinious
But I am not sure about 90 and 270. Tangent is undefined at these points.
According to the book I am wrong but Ahri is right.
f:A--->R
a is element of A
There is a limit when x goes to a and this limit is equal to f(a).
If we put our finger on one end of the function and go to the other end without removing our finger then we can say that the function is continious.