Question

Verify Rolle’s theorem for the function f where f(x) = sin x, x Î [-2π, 2π].

Answer 1

Continuous in \([-2\pi, 2\pi]\)? Yes.
Differentiable in \((-2\pi, 2\pi)\)? Yes.
\(f(2\pi) = f(-2\pi)\)? Indeed.

Answer 2

So when exactly is the slope of the tangent zero?\[f'(x) = \cos (x) = 0 \Rightarrow x=\pm \pi/4, \pm 3\pi /4 \]

Answer 3

OK. Anything else Sir?

Answer 4

That's very much it.

Answer 5

OK. Thanks

Answer 6

f'(x)=cosx=0 x=+-pi/4

Answer 7

,+-3pi/4

Answer 8

please can you explain that line for me ?

Answer 9

@perl

Answer 10

cos x = 0 when x = pi/2 , 3pi/2 , 5pi/2, etc

Answer 11

thanks bosssssssssssssssssssssssssssssssssss

Answer 12

I think parth made a typo above

Answer 13

yes. thanks for the correction

Answer 14

yap. i got that.

Answer 15

Continuous in \( [-2\pi, 2\pi] \) ? Yes.
Differentiable in \((-2\pi, 2\pi)\)? Yes.
\(f(2\pi) = f(-2\pi)\)? Indeed.

So when exactly is the slope of the tangent zero?\[f'(x) = \cos (x) = 0 \Rightarrow x=\pm \pi/\color{red}{2}, \pm 3\pi /\color{red}{2}\]

Answer 16

perl, is there a shorter way to prove continuity and derivability ?

Answer 17

well, sin x and cos x are continuous and differentiable for all real numbers

Answer 18

you don't need to prove that here. we can assume that it has been proved earlier , and it is a theorem

Answer 19

Continuous in \( [-2\pi, 2\pi] \) ? Yes because it is continuous for all reals.
Differentiable in \((-2\pi, 2\pi)\)? Yes. because it is differentiable for all reals.
\(f(2\pi) = f(-2\pi)\)? Indeed.

So when exactly is the slope of the tangent zero?\[f'(x) = \cos (x) = 0 \Rightarrow x=\pm \pi/\color{red}{2}, \pm 3\pi /\color{red}{2}\]

Answer 20

ok

Answer 21

can you prove this if continous and derivable ?
\[\left| x \right| +\left| x+1 \right| for x=0 and x=1\]

Answer 22

can you prove this if continuous and derivable ?
\[ \Large {\left| x \right| +\left| x+1 \right| ~for~ x=0,1}\]

Answer 23

yes

Answer 24

\( \large f(x) = |x | \) is continuous and \( \large g(x) = |x+1 | \) is continuous, and the sum of two continuous functions is continuous, so \( \large f(x) + g(x) \) is continuous .

Answer 25

ok

Answer 26

now for the differentiable part, lets define the absolute value function in terms of cases , or piecewise define it

Answer 27

waw.

Answer 28

$$ \Large{
|x| + |x+1|=
\begin{cases}
x + (x+1) & \text{if } x \geq 0\\
-x + (x+1) & \text{if } -1\leq x<0\\
-x - (x+1)& \text{if } x<-1
\end{cases}}
$$ simplified this gives us

$$\Large{
|x| + |x+1|=
\begin{cases}
2x+1 & \text{if } x \geq 0\\
1 & \text{if } -1\leq x<0\\
-2x - 1& \text{if } x<-1
\end{cases}
}


$$

Answer 29

pk

Answer 30

ok got that

Answer 31

$$
\Large{
|x| + |x+1|=
\begin{cases}
2x+1 & \text{if } x \geq 0\\
1 & \text{if } -1\leq x<0\\
-2x - 1& \text{if } x<-1
\end{cases}
} \\{}
\\
\Large{
(|x| + |x+1|) '=
\begin{cases}
2 & \text{if } x > 0\\
undefined & \text{if } x = 0\\
0 & \text{if } -1< x<0\\
undefined & \text{if } x = -1\\
-2 & \text{if } x<-1
\end{cases}
}



$$

Answer 32

do you see why the derivative is undefined at x = 0 and x = -1, because the slope changes depending on which direction you are coming from, from the right or left of x = 0 , right or left of x = -1. The slopes do not match at the point .

Answer 33

waw. this is great

Answer 34

are you a lecturer ?