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### Course: AP®︎ Calculus AB (2017 edition)>Unit 5

Lesson 2: Mean value theorem

# Conditions for MVT: graph

The mean value theorem applies to a function ƒ over an interval [𝘢,𝘣] under the conditions that ƒ is differentiable over (𝘢,𝘣) and continuous over [𝘢,𝘣]. See how we determine these conditions given a graph.

## Want to join the conversation?

• At , why is it differentiable over the open interval? And also, let a function which has an end point at c. Is the function continous at c?
• Why is the graph `h` differentiable over the open interval `[-3, 2]`?
I believe differentiable pretty much means that if someone asked you "What is the slope of the line at spot?" you would know the answer, unlike at .

So the function h is differentiable over the open interval `[-3, 2]` because I can tell you without a doubt that the slope at any point in that interval is 1.

Let a function which has an end point at C. Is the function continuous at C? Continuous means there are no breaks in the function's graph, unlike in the video above at .
The answer to your second question depends on what the graph looks like at C.

Hope this helps, and feel free to ask more questions!
(1 vote)
• Does a function need to be continuous/differentiable to have one tangent line? Secant line?
(1 vote)
• Find the number c that satisfies the mean value theorem for the function ((4/x)+x) on the interval [1,4]
• 𝑓(𝑥) = 4 ∕ 𝑥 + 𝑥 is differentiable over the interval [1, 4], so the mean value theorem is applicable.

This means that there exists a 𝑐 ∈ [1, 4] for which 𝑓 '(𝑐) is equal to the slope of the straight line between the points (1, 𝑓(1)) and (4, 𝑓(4)).

First of all, the slope is (𝑓(4) − 𝑓(1)) ∕ (4 − 1) = (4 ∕ 4 + 4 − (4 ∕ 1 + 1)) ∕ 3 = 0.

Secondly, 𝑓 '(𝑥) = −4 ∕ 𝑥² + 1 ⇔ 𝑓 '(𝑐) = −4 ∕ 𝑐² + 1

So, we want to solve the equation −4 ∕ 𝑐² + 1 = 0 ⇔ 𝑐² = 4 ⇔ 𝑐 = ±2
However, 𝑐 ∈ [1, 4] ⇒ 𝑐 = 2