Question

Does this equation has a solution at intervals [-2,10] ?

Answer 1

\[2^{-x ^{2}}-x=0\]

Answer 2

So if we add some stuff, then multiply some stuff over, we get,
\[\Large\bf 2^{-x^2}\quad=\quad x\qquad\qquad\to\qquad\qquad 1\quad=\quad x\;2^{x^2}\]

So maybe it's a little easier to work with from here.
We want to know if \(\Large \bf y=1\) intercepts the function \(\Large\bf y=x\;2^{x^2}\)

I know we can probably use some calculus techniques to show that if x 2^{x^2} is continuous and is greater than and less than 1 at different points in the interval, then we can probably justify that it has to pass through the line y=1.

I'm not sure how we would find the `actual` solution though.. hmm
Have you done problems like this before?
Is there a special technique? :o

Answer 3

hm, i do not have the solution, its just says if the a solution can be obtained by using intermediate value theorem. The answer is yes. but how am i suppose to prove that there is indeed a solution without calculating it, if the solution is that complicated.

Answer 4

how come you cannot apply intermediate value theorem in equations like,
1/x=0,
x^2-7x+10=0,
x/(x^2-1)=0?

Answer 5

Well the IVT requires that the function be `continuous` over the given interval.
So for the first one:
1/x=0, we would NOT want to apply the IVT to an interval which includes x=0, right?

Answer 6

Since there is a discontinuity there*

Answer 7

hmm okay so you're saying, as long as x doesnt equal to 0? so that means the first equation how do u know there is a solution at [-2,10]

Answer 8

So like if we can show that:

f(-2) = (less than 1)
`and`
f(10) = (greater than 1)
`and`
the function is continuous

`then`
the function MUST HAVE passed through y=1 somewhere in the interval.

Answer 9

So we're talking about something like this.
If we can find points like this in our interval,