
The Mean Value Theorem (MVT) connects the average rate of change of a function to its instantaneous rate of change at some point within an interval.
In simple terms:
Somewhere between two points on a smooth curve, there is at least one point where the tangent line is parallel to the secant line joining those two points.
Let $$f(x)$$ be a function that satisfies two conditions:
$$f(x)$$ is continuous on the closed interval [a,b].
$$f(x)$$ is differentiable on the open interval (a,b).
Then, there exists at least one number c∈(a,b) such that:
$$f^\prime(x) = \frac{f(b) – f(a)}{b – a}$$
The right-hand side, $$\frac{f(b) – f(a)}{b – a}$$, represents the average rate of change of $$f(x)$$ over [a,b].
The left-hand side, $$f^\prime(x)$$, represents the instantaneous rate of change (slope of the tangent) at x=c.
Thus, MVT says:
There exists a point ccc where the instantaneous rate of change equals the average rate of change.
Imagine a smooth curve from point A(a , f(a)) to B(b , f(b)).
The secant line connects A and B.
The tangent line at some point C(c , f(c)) is parallel to that secant line.
That’s the Mean Value Theorem in action!
Find the point ccc that satisfies the Mean Value Theorem for:
$$f(x) = x^{2} on [1,3]$$
Step 1. Check the conditions:
$$f(x) = x^{2}$$ is continuous and differentiable everywhere.
Step 2. Compute the average rate of change:
$$\frac{f(3) – f(1)}{3 – 1} = \frac{9-1}{2} = 4$$
Step 3. Find c such that $$f^\prime(c) = 4$$.
$$f^\prime(x) = 2x$$ then $$2c = 4 \longrightarrow c=2$$
at c=2, the tangent line is parallel to the secant line between x=1 and x=3.
Let $$f(x) = \sin(x)$$ on $$[0,\pi]$$.
$$\frac{f(\pi) – f(0)}{\pi – 0} = \frac{0}{\pi} = 0$$
Find c such that $$f^\prime(c)=0$$.
$$f^\prime(x) = \cos(x)$$ then $$\cos(c) = 0 \longrightarrow c = \frac{\pi}{2}$$
$$c = \frac{\pi}{2}$$satisfies the Mean Value Theorem.
You have not completed all required lessons and assessments.