Over what interval will the immediate value theorem apply

Mathematics

1 answer:

koban [17]3 years ago

6 0

Answer:

Any [a,b] that does NOT include the x-value 3 in it.

Either an [a,b] entirely to the left of 3, or

an [a,b] entirely to the right of 3

Step-by-step explanation:

The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.

Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.

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Step-by-step explanation:

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Take a look at the picture attached. It has a "formula": ∠a+∠b+∠c=180°

Use this as it will help you a lot. You plug in the info you already know (it doesn't matter which angle is a,b,c) and solve for the missing one. :)

Ask in the comments below if you are not sure about something.

I hope that helps you! :D