How can I find the interval where the zero exists?

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lmxz9

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If I got it right then the zero exists between x=0 and x=1. So for y=0 its x must be a very weird number. I really have no idea how to calculate it. Please, help!
 
lmxz9 said:
View attachment 24681
If I got it right then the zero exists between x=0 and x=1. So for y=0 its x must be a very weird number. I really have no idea how to calculate it. Please, help!
Click to expand...
Why do you think it's weird? It could be 0.376. Is it a weird number?
What method did you use for similar problems in class? Any examples in your book?
 
You're looking for an x-axis crossing, i.e.

f(lo) has a different sign than f(hi)

You're just going to have to plug these all in and check the signs.

Clipboard01.jpg

from that which interval appears to have a sign change within it?
 
lmxz9 said:
View attachment 24681
If I got it right then the zero exists between x=0 and x=1. So for y=0 its x must be a very weird number. I really have no idea how to calculate it. Please, help!
Click to expand...
You don't need to actually calculate this value of x; you just need to decide which interval contains it. In fact, this is something you could use to find an approximate solution to an equation when it is impossible to find an exact value.

The trick you have presumably been taught is the Intermediate Value Theorem:


Read that; it has a very nice explanation of the idea and how to use it.
 
lev888 said:
Why do you think it's weird? It could be 0.376. Is it a weird number?
What method did you use for similar problems in class? Any examples in your book?
Click to expand...
Just the standard methods: factorise or find and divide the factors of the constant over the factors of the leading coefficient. This kind of exercise is not in the book we use (Precalculus by Stewart) but was part of last year's final exam T^T

Romsek said:
You're looking for an x-axis crossing, i.e.

f(lo) has a different sign than f(hi)

You're just going to have to plug these all in and check the signs.

View attachment 24682

from that which interval appears to have a sign change within it?
Click to expand...
Oh! So, is it because x^3+2x-1 changes signs once?

Dr.Peterson said:
You don't need to actually calculate this value of x; you just need to decide which interval contains it. In fact, this is something you could use to find an approximate solution to an equation when it is impossible to find an exact value.

The trick you have presumably been taught is the Intermediate Value Theorem:


Read that; it has a very nice explanation of the idea and how to use it.
Click to expand...
Aaaaa!!! The IVT! First time I hear about it to be honest. So in this case my "c" was 0. I had to try the answers one by one. Like in b) f(0.4)=-0.136 and f(0.5)=0.125, the number 0 (c) is between them, so the zero is indeed between 0.4 and 0.5! The other options didn't satisfy this. Thank you so much!!! I now understand everything! :D P.S: That page was so helpful! ^^

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I really appreciate your comments, everyone! Thank you so much, you all took the time to try to help me! I'm so grateful! Wish you a beautiful beautiful day! <(^.^)>
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Have you ever purchased something for $0.85? What a weird price?
 
lmxz9 said:
Aaaaa!!! The IVT! First time I hear about it to be honest. So in this case my "c" was 0. I had to try the answers one by one. Like in b) f(0.4)=-0.136 and f(0.5)=0.125, the number 0 (c) is between them, so the zero is indeed between 0.4 and 0.5! The other options didn't satisfy this. Thank you so much!!! I now understand everything! :D P.S: That page was so helpful! ^^
Click to expand...
Excellent. And the zero turns out to be approximately 0.453, according to a graph.

But it's odd that you would be assigned this problem without having been taught anything about the IVT, or that your book wouldn't even mention it.
 
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