Using the Mean Value Theorem
Introduced| English |
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| Mean Value Theorem/miːn ˈvæljuː ˈθɪərəm/ |
Somewhere, the instant matches the average
- Drive $120$ km in $2$ hours: your average speed is $60\,\tfrac{\text{km}}{\text{h}}$.
- At some instant your speedometer must have read exactly $60$ — you can't average $60$ without hitting it.
- That intuitive fact is the Mean Value Theorem 中值定理 (MVT).
- It links the average rate of change over an interval to an instantaneous rate somewhere inside.
The hypotheses come first
- MVT needs $f$ continuous on $[a,b]$ and differentiable on $(a,b)$.
- If both hold, there is at least one $c$ in $(a,b)$ with
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$$f'(c)=\frac{f(b)-f(a)}{b-a}$$
- The left side is an instantaneous rate; the right side is the average rate over $[a,b]$.
The MVT on $[a,b]$ requires which hypotheses?
Continuous on the closed interval and differentiable on the open one. ($f(a)=f(b)$ is the extra Rolle condition.)
The MVT guarantees a $c$ where $f'(c)$ equals the...
$f'(c)=\frac{f(b)-f(a)}{b-a}$, the average rate.
A tangent parallel to the secant
- The right-hand side is the slope of the secant line joining the endpoints.
- So MVT guarantees a point where the tangent line is parallel to that secant.
- Geometrically: somewhere the curve's slope equals the overall average slope.
- Like the IVT, it is an existence theorem — it promises $c$ exists, not its value.
Tangent parallel to the secant
y = x²
Somewhere inside the interval the tangent slope matches the average (secant) slope — that point is the MVT's guaranteed $c$.
Geometrically, the MVT guarantees a tangent line parallel to the ____ line.
The tangent at $c$ is parallel to the secant joining the endpoints.
Finding the guaranteed $c$
- To find $c$: set $f'(c)$ equal to the average rate and solve for $c$ in $(a,b)$.
- First check the hypotheses (continuous + differentiable) — skip that and the theorem doesn't apply.
- Discard any solution outside the open interval $(a,b)$.
- The MVT is the engine behind many later results (like "if $f'=0$ everywhere, $f$ is constant").
For $f(x)=x^2$ on $[2,6]$, find the $c$ guaranteed by the MVT.
Average $=\frac{36-4}{4}=8$; $2c=8\Rightarrow c=4\in(2,6)$.
If $f$ has a corner inside $(a,b)$, the MVT is still guaranteed to apply.
A corner breaks differentiability, so the hypotheses fail and MVT may not apply.
Like the IVT, the MVT is what kind of theorem?
It guarantees such a $c$ exists (though here you can often find it).
Both hypotheses are required. If $f$ has a corner (not differentiable) or a break (not continuous) on the interval, the MVT can fail — there may be no such $c$. Always confirm $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ before applying it.
For $f(x)=x^2$ on $[1,3]$, find the $c$ guaranteed by the MVT.
- Average rate: $\dfrac{f(3)-f(1)}{3-1}=\dfrac{9-1}{2}=4$.
- $f'(x)=2x$, so set $2c=4\Rightarrow c=2$.
- $c=2$ lies in $(1,3)$ ✓ — there the tangent slope equals the average slope $4$.
The Mean Value Theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, some $c\in(a,b)$ has $f'(c)=\frac{f(b)-f(a)}{b-a}$ — an instantaneous rate equal to the average rate, i.e. a tangent parallel to the secant. Check the hypotheses, then solve $f'(c)=$ average for $c$.