What is the mean value theorem?
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
The MVT defines a point in an interval where the slope of the tangent line equals the slope of the secant line, by using this formula:
\begin{equation}
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}
\end{equation}
But how?
Let’s dive in further to see exactly how and why this process works.
The Backstory
Remember our previous lesson on Rolle’s Theorem?
Well, Rolle’s Theorem is a special case of the Mean Value theorem, and is critical in the proof of MVT as it considers functions where endpoints don’t necessarily yield the same value, but where we can find a point where two slopes are equal.
What two slopes? Slope of the tangent line and the slope of the secant line.
Instantaneous Vs Average
The slope of the tangent line is the instantaneous rate of change at a point on a curve.
To calculate the instantaneous rate of change, we use our derivative rules and substitute a value into the first derivative to find the slope at a point.
In algebra, we learn that the slope of a secant line represents the average rate of change over time. To find the average rate of change, we apply the slope formula.
Tangent Line Vs Secant Line
Mean Value Theorem
The Big Idea
So the Mean Value Theorem (MVT) allows us to determine a point within the interval where both the slope of the tangent and secant lines are equal.
Now, let’s think geometrically for a second. If two linear are parallel, then we know that they have the same slope. This means we are on the hunt for parallel lines.
Moreover, if we know that parallel lines have the same slope, then all we have to do is set the slope of one line equal to the other and solve!
MVT: Tangent Equals Secant
And this is the idea behind the MVT.
Formula
The Mean Value Theorem states that if f is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in the open interval (a,b) such that:
\begin{equation}
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}
\end{equation}
MVT Example – How To Find C?
For example, suppose we are given \(f(x)=x^{2}\) on the interval [-2,1], and we want to find all values of c in the open interval (-2,1) such that the instantaneous rate of change equals the average rate of change.
Step 1:
First, we must verify that the function f(x) is continuous on the closed interval and differentiable on the open interval. Because f(x) is a polynomial function, we are guaranteed that f(x) is both continuous and differentiable.
Step 2:
Next, we calculate the instantaneous rate of change by using the power rule and get:
\begin{equation}
f^{\prime}(x)=2 x
\end{equation}
Step 3:
Now we calculate the average rate of change on the closed interval [-2,1] using our slope formula.
\begin{equation}
\frac{f(1)-f(-2)}{1-(-2)}=\frac{(1)^{2}-(-2)^{2}}{1+2}=\frac{-3}{3}=-1
\end{equation}
Step 4:
Finally, we set our instantaneous slope equal to our average slope and solve.
\begin{equation}
\begin{array}{l}
2 x=-1 \\
x=\frac{-1}{2} \\
c=\frac{-1}{2}
\end{array}
\end{equation}
Therefore, we have found that in the open interval c = -1/2, which means at this location, the slope of the tangent line equals the slope of the secant line.
Apply Mean Value Theorem Example
In this video, we will discover when the MVT does and doesn’t apply and find all values of c within the open interval that satisfy the conclusion of this remarkable theorem!
Let’s get to it!
Video Tutorial w/ Full Lesson & Detailed Examples (Video)
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