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Mean Value Theorem (MVT) Calculator

Find every c in (a, b) where the instantaneous rate equals the average rate of change.

Enter your values

ⓘ Live results

Results
Average rate of change
1
f(a)
-2
f(b)
2
c value(s) in (a, b)
-1.154701, 1.154701

At each c the tangent line is parallel to the secant through (a, f(a)) and (b, f(b)).

Working

  1. 1Secant slope = (f(2) − f(-2)) / (2 − -2) = 1
  2. 2Solve f′(c) = secant slope for c inside the open interval.
  3. 3c = -1.154701, 1.154701
x = -2f(x) = x^3 - 3xx = 2

Quick answer

The mean value theorem says that for a function continuous on [a, b] and differentiable on (a, b) there is at least one c where f′(c) = (f(b) − f(a)) / (b − a). This calculator finds every such c.

f′(c) = [f(b) − f(a)] / (b − a)

How to use the Mean Value Theorem Calculator

  1. 1Enter a function that is continuous on the closed interval.
  2. 2Enter the endpoints a and b.
  3. 3Compare the secant slope with f′(c) at each returned c.

Worked example

Using the values the calculator loads with:

Inputs

  • Function f(x)x^3 - 3x
  • Interval start a-2
  • Interval end b2

Results

  • Average rate of change1
  • f(a)-2
  • f(b)2
  • c value(s) in (a, b)-1.154701, 1.154701
  1. Secant slope = (f(2) − f(-2)) / (2 − -2) = 1
  2. Solve f′(c) = secant slope for c inside the open interval.
  3. c = -1.154701, 1.154701

Frequently asked questions

What are the hypotheses of the mean value theorem?

f must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). If either fails, the conclusion can fail too.

Can there be more than one c?

Yes. The theorem guarantees at least one c, but many functions have several points where the tangent is parallel to the secant.

How is the MVT related to Rolle's theorem?

Rolle's theorem is the special case where f(a) = f(b), which makes the secant slope zero, so f′(c) = 0.

About this calculator

Results are estimates for general information. For medical, legal, structural or financial decisions, confirm with a qualified professional.