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### Course: Differential Calculus>Unit 1

Lesson 16: Intermediate value theorem

# Worked example: using the intermediate value theorem

Discover how the Intermediate Value Theorem guarantees specific outcomes for continuous functions. With a given function f, where f(-2) = 3 and f(1) = 6, learn to identify the correct statement that aligns with the theorem's principles and understand its real-life applications.

## Want to join the conversation?

• I didn't quite understand why can't the function take up values that are higher or lower than the boundary values i.e. the values of f(b) and f(a)?
If the value for f(a)=5 and that for f(b)=-3 then the value of f(c) has got to be between -3 and 5, why is it so? Why Can't it be like f(c)=6 or may be f(c)= -7, where c lies between a and b ?
• The theorem says that the function achieves every value between f(a) and f(b). It does not say that it cannot reach other values as well. This is perfectly possible, as you observe.
• I don't understand. If y<3 or y>6 and it's still a continous function, then how could the theorem guarantee?
• In this example, the theorem only guarantees that we will have all values that are greater than 3 and less than 6 (such as 4, here). We could get a y that is less than 3 or greater than 6, but that isn't guaranteed by continuity.
• Isn't the intermediate value theorem just basic common sense or is there more to it?
• It certainly feels like common sense, especially with our intuition about graphs of functions. But a truly rigorous proof of it requires a bit of work. "Common sense" is often wrong in math, which is why we must write careful proofs, even of simple-sounding statements.
• So if the intermediate value theorem is not followed by a function in a given interval, can we say that it's not continuous?
• Yes, by contraposition. All continuous functions have the intermediate value property, so anything without the intermediate value property can't be continuous.
• Why is the second option ruled out ? why can't f(c) be equal to 0? A function can be anything . Imagine drawing a sort of parabola which drops down from 3 to 0 and then reaches 6. why cant that be not possible?
• The function could be zero, but it's not guaranteed to be. We're looking at what values the function must achieve on the given interval.
• About a question on the cotinutity challenge in the next lesson, I still don't get understand why is ^3√x+1 defined for all real numbers and ^4√x+1 is not? I think both are defined for all x values such that x+1≧0.
• The issue here is that negative numbers can have odd-numbered roots, but not even-numbered roots. The cube root of -1 is -1, because (-1)*(-1)*(-1) = -1. However, you can't take the fourth root of a negative number, just as you can't take its square root (usually).

In the problem you're referencing, I at least was given an x value of -2. If your question wasn't different, then we're looking for the cube root and fourth root of (x+1). Since 1 - 2 = -1, the cube root is defined for x = -2, but the fourth root is not. Since the value is not defined, the function is not continuous at x = -2.
• what is the purpose of intermediate value theorem?
• The intermediate value theorem can give information about the zeros (roots) of a continuous function. If, for a continuous function f, real values a and b are found such that f(a) > 0 and f(b) < 0 (or f(a) < 0 and f(b) > 0), then the function has at least one zero between a and b.

Have a blessed, wonderful day!
• Why can't the function go lower than 3 or higher than 6?
• Actually, it is very possible for the function to exceed those values in either direction, especially beyond the concerned interval. The IVT only tells us that for this case, every value between 3 and 6 is represented at some point of the function.