Example 1 — \(f(x)=x^2\) on \([1,3]\)
\(m_{\text{sec}}=\dfrac{9-1}{3-1}=4\). Solve \(f'(x)=2x=4\Rightarrow c=2\in(1,3)\). The tangent at \(x=2\) has slope \(4\), matching the secant.
Find points \(c \in (a,b)\) such that \(f'(c)=\dfrac{f(b)-f(a)}{b-a}\). Enter your function and interval, then press Calculate.
A Mean Value Theorem (MVT) Calculator analyzes a function on a closed interval and pinpoints the location(s) where the instantaneous rate of change equals the average rate of change. For a function continuous on \([a,b]\) and differentiable on \((a,b)\), the MVT guarantees a number \(c\in(a,b)\) such that the tangent slope at \(c\) equals the secant slope across the interval. The calculator verifies the hypotheses automatically, computes the secant slope, solves \(f'(x)=\frac{f(b)-f(a)}{b-a}\) for \(x\in(a,b)\), filters valid solutions, and shows each algebraic step. It can also illustrate Rolle’s Theorem as a special case and provide the generalized Cauchy MVT when two functions are supplied.
The tool accepts symbolic functions (polynomials, exponentials, logs, trig, rationals) or numeric black-box functions. It checks continuity on \([a,b]\) and differentiability on \((a,b)\) via domain tests and derivative existence; if assumptions fail, it flags the issue and explains why the theorem may not apply. When multiple points satisfy the condition, all solutions in \((a,b)\) are listed. For constant functions, any point in \((a,b)\) works; the calculator reports the entire interval or a representative set. For two functions \(f\) and \(g\) with \(g'\neq0\), it computes the Cauchy Mean Value Theorem ratio. Optional numeric sampling illustrates how the tangent at \(c\) parallels the secant from \((a,f(a))\) to \((b,f(b))\).
Average (secant) slope: \[ m_{\text{sec}}=\frac{f(b)-f(a)}{\,b-a\,}. \]
Mean Value Theorem: \[ \text{If } f\in C[a,b] \text{ and } f\in C^1(a,b),\ \exists\,c\in(a,b):\quad f'(c)=\frac{f(b)-f(a)}{b-a}. \]
Rolle’s Theorem (special case): \[ f(a)=f(b),\ f\in C[a,b],\ f\in C^1(a,b)\ \Rightarrow\ \exists\,c\in(a,b):\ f'(c)=0. \]
Cauchy Mean Value Theorem: \[ \exists\,c\in(a,b):\ \frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)},\ \text{with } g'(x)\ne0 \text{ on }(a,b). \]
\(m_{\text{sec}}=\dfrac{9-1}{3-1}=4\). Solve \(f'(x)=2x=4\Rightarrow c=2\in(1,3)\). The tangent at \(x=2\) has slope \(4\), matching the secant.
\(m_{\text{sec}}=\dfrac{0-0}{\pi-0}=0\). Solve \(f'(x)=\cos x=0\Rightarrow c=\tfrac{\pi}{2}\in(0,\pi)\).
\(m_{\text{sec}}=\dfrac{1-1}{2}=0\). But \(f'(x)=-1\) for \(x<0\), \(f'(x)=1\) for \(x>0\); no \(c\) with \(f'(c)=0\). The MVT hypotheses fail because \(f\) is not differentiable at \(0\).
Continuity on \([a,b]\) and differentiability on \((a,b)\). Without both, the guarantee may fail.
Yes. The equation \(f'(x)=\dfrac{f(b)-f(a)}{b-a}\) can have several solutions inside \((a,b)\).
Rolle’s is the special case \(f(a)=f(b)\), guaranteeing a point with zero derivative.
It explains which condition breaks and shows why a point \(c\) may not exist (e.g., corners, cusps, discontinuities).
Swap endpoints to enforce \(a<b\); slopes and conclusions are unchanged.
Yes. With \(f\) and \(g\), it applies the Cauchy Mean Value Theorem provided \(g'(x)\neq0\) on \((a,b)\).