Intermediate value theorem to prove a root in an interval (KristaKingMath)

Learn how to use the intermediate value theorem to prove that a function has a root (point of intersection with the x-axis) on a given interval. To prove that the root exists in the interval, plug the left and right endpoints of the interval into the function. If the left endpoint produces a negative value and the right endpoint produces a positive value, then you can use the intermediate value theorem to prove that the graph crosses the x-axis (has a value of zero) somewhere between the endpoints.

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Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)

Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”

[zachz]

Hey, this video is 8 years old but you just saved my ass. I wish you the best.

[darkmusky]

Short, easy explanation and straight to the point YOU SAVED ME ON MY EXAM

[mdreadking]

I found your presentation an explanation to be simple and clear. To have it explained in the manner presented I found made very clear the text book math verbiage. The reason we're all taking Calculus is because we haven't. I believe one of the keys to understanding exact science is understating or interpreting what is being asked.
Great job, Thanks

[xgirlie]

MY PROF TOOK TWO HOURS TO EXPLAIN THIS AND I DIDNT UNDERSTAND A THING. YOU DID IT IN 5 MINS AND IT MAKES SENSE. THANK YOU SO MUCH

[kristakingmath]

I was a Psychology major. I'm so glad the videos are helping you!! :)

[kristakingmath]

I was a calculus tutor all four years in college. Had I realized early on, I would have likely been an engineer or math major. :)

[kristakingmath]

by definition, the "root" of a function is where it crosses the x-axis, and at the x-axis, the function's value is 0. :)

[alitr-tfdc]

even after 12 years we are still making use of it, thanks alot💙

[dontjudgeme]

gotta love tutors who use Stewart's book ! i was reviewing a little bit before starting the new semester and got stuck at this question ^^ luckily here you are explaining the theorem with the exact question !! thanks ^^

[PoodleKing]

Those coordinate axes are kinda confusing. If you want to show there's a root you should have the graph you draw to illustrate it cross the x-axis just for clarity's sake.

[Evelyn-grmp]

It is so nice to see a woman teaching teaching a helpful math video! -from another fellow woman stem major. keep up the great work!

[Kosovathereality]

Just discovered your awesome math channel, you have compliments for your clear way of teaching!

[kristakingmath]

yes, you assume that f(c)=0 every time, because you're always looking for the point at which the graph crosses the x-axis, and that can only happen when f(c)=0. i hope that helps! :)

[LAWRENCExRACHAL]

Thank you so much for taking the time to explain this theorem. This is one of the most clear and concise math explanations I've ever come across. I only had to watch one time and I understood. Thank you.

[kwames.]

your voice is so soft and sweet...A refreshing step up from my old senile prof. Awesome Vid.

[kristakingmath]

no, you just need to show that the function is negative at some point, and positive at some point. because if you can show that it's below the x-axis at some point, and above the x-axis at another point, then you know it has to cross the x-axis somewhere, and you've proven that it has a solution. :)

[kristakingmath]

if you can take the derivative of the function, set that equal to 0, and solve for x, and if you get two x-values as a result that are both inside your interval, then you know that the function has two critical points in the interval, which means it changes direction twice. if you can find the value of the function at the endpoints, and the value at the critical points, then you can show that it moves back and forth across the x-axis. :)

[kristakingmath]

Awesome! I'm so glad I'm able to help along the way. :)

[kristakingmath]

Sorry it's quiet! I'll try to make it louder. :)

[Crynuslord]

The theorem only holds true if the function is continuous on the closed interval correct??? I didn't hear you mention that... maybe I missed it.