MATH 136
The Mean Value Theorem
Let f be a continuous function on the interval [ a , b ] such that f ′(x) exists on ( a , b ).
The Mean Value Theorem (MVT) states that there is a point in ( a , b ) where the exact rate
of change equals the average rate of change. That is,
f (b) − f (a)€ €
has a solution
f ′(x) =
€ €in ( a , b ).
b−a
€ €
Example 1. Apply the MVT to f ( x) = x 2 − 6x + 9 on [2, 6].
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Solution. Because f is a continuous polynomial that differentiable everywhere, the
MVT applies. The average rate of change of f over [2, 6] is given by
c=
f (6) − f (2) 9 − 1
=
= 2,
6−2
4
while the exact rate of change is f ′ (x ) = 2x − 6 . The MVT states that f ′ (x ) = c has a
solution in the interval (2, 6). Thus, we solve 2x − 6 = 2 , to obtain x = 4 . Hence, when
x = 4 , the exact rate of change f ′ (4) is equal to the average rate of change of 2.
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€ can use the solve command
If we do not wish to solve f ′ (x ) = €
c “by hand,” then we
on a calculator after evaluating c = ( f (b) − f (a)) / (b − a) .
TI-84 Method 1: Enter f ( x) into Y1. If f ′(x) is not to difficult to compute, then take
the derivative and enter it into Y2. Evaluate c = (Y1(b)–Y1(a))/(b–a). Then enter the
command solve(Y2–c, €
X, guess), where a good initial guess is usually the midpoint of
the interval ( a , b ). (On the TI–89, use solve(y2(x)=c, x = guess).)
TI-84 Method 2: If you do not want€to take the derivative, then enter f into Y1, and
enter the numerical derivative into Y2: Y2 = nDeriv(Y1, X, X). (On the TI–89, use
€ €x) .) Then enter the command solve(Y2–c, X, guess).
∂ (y1(x),
Example 2. Use numerical methods on a calculator to apply the MVT to f (x) = x 2 sin x
on [ − π / 2, π ] .
Solution. The MVT applies because f is continuous and differentiable everywhere. We
€
now enter x 2 sin x into Y1, and enter nDeriv(Y1, X, X) into Y2. To make the evaluations
easier, we store –π/2 to A, store π to B, then enter and store the average rate of change
(Y1(b)–Y1(a))/(b–a) to C. Finally enter solve(Y2–C, X, π/4).
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We obtain x ≈ 0.428714673 . That is, f ′ (0.428714673) ≈ 0. 5235987756 ≈ c .
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Note: If we use the true derivative function Y3 = 2Xsin(X) + X2 cos(X) , then we obtain
a more precise solution of x ≈ 0.4287149769 .
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