Question

Does the function has a zero in the given interval? Explain why.

This is a question on IVT
Intermediate value theorem

Answer 1

f(x) = tan(2x) - 4 on \([-\pi, \pi]\)

\(\Large f(-\pi) = tan(2( -\pi)) = \frac{sin(-2\pi)}{cos(-2\pi)}=\frac{-sin(2\pi)}{cos(2\pi)} =0\)

\(\Large f(\pi) = tan(2( \pi)) = \frac{sin(2\pi)}{cos(2\pi)} =0\)

Answer 2

lol there is no like picture?

Answer 3

nvm

Answer 4

f(x) = tan(2x) - 4 on \([-\pi, \pi]\)

\(\large f(-\pi) = tan(2( -\pi))-4 = \frac{sin(-2\pi)}{cos(-2\pi)}-4=0 - 4 =-4\)

\(\large f(\pi) = tan(2( \pi))-4 = \frac{sin(2\pi)}{cos(2\pi)} -4=0 - 4 = -4\)

Answer 5

The function does have zeroes in the given interval (you can graph it to see the zeroes).
But the Intermediate Value Theorem can not be applied since the function is not continuous on the given interval. So the IVT can not guarantee us that a zero exists in the interval.

Answer 6

So by IVT, since a and b are equal, the function isn't continuous and we won't know if there is a 0 in between the range unless we use a smaller range that is continuous

Answer 7

Oh, it fails for two different reasons, yes good call :)
We don't have a<0 and b>0 like you said.

Answer 8

Yayay, *high five* We got the same solution at the same time xDD

And I did graph it haha https://www.desmos.com/calculator/qoapaubntx

Answer 9

Is yours graphed properly?
Did I do mine incorrectly?
https://www.desmos.com/calculator/4y1wt1oras
Hmm

Answer 10

Oh I see, you skewed the y-axis.

Answer 11

I didn't know you can do the notation like that :d interesting

Answer 12

Yeah, my y-axis has a huge range apparently xD

Answer 13

And likewise, I find yours much more cleaner to use

I had to search it up on desmos on how to add a restriction and that's the way they showed so I've been using that for the past week :3

Answer 14

\(\color{#0cbb34}{\text{Originally Posted by}}\) @TheSmartOne
So by IVT, since a and b are equal, the function isn't continuous and we won't know if there is a 0 in between the range unless we use a smaller range that is continuous
\(\color{#0cbb34}{\text{End of Quote}}\)

want to correct a little mistake I made

it isn't continuous because f(a) and f(b) are equal