# Mean Value Theorem

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Mean value theorem (MVT), it is also named as Lagrange's Mean Value Theorem.
Statement : Let f(x) be a function defined on [a,b] such that
1) it is continuous on [a,b].
2) it is differentiable on (a,b).
Then, there exists a real number c $\epsilon$ (a,b) such that

$f'(c) = \frac{f(b)-f(a)}{b - a}$

Geometrical representation of the mean value theorem
Let f(x) be a function defined on [a,b] and let APB be a curve represented by y = f(x). Then coordinates of A and B are (a,f(a)) and (b,f(b)) respectively and P (c,f(c)) Suppose the chord AB makes ant angle $\theta$ with the x-axis. Then we got a triangle as ARB and so we have ,
tan $\theta = \frac{BR}{AR}$
$\Rightarrow tan \theta = \frac{f(b) -f(a)}{b - a }$ Slope of chord AB = slope of the tangent at P(c,f(c)).
Slope of tangent = tan $\theta$ = f '(c)

tan $\theta = \frac{f(b) -f(a)}{b - a }$

f '(c) = $\frac{f(b) -f(a)}{b - a }$

## Examples on Mean Value Theorem

Example 1 : Verify the MVT for the function f(x) = $x^{3} -18x^{2} + 99x -162$ on the interval [3,5].
Solution : As the given function is a polynomial so it is continuous and differentiable in the given interval. So there must exist at least one real number c $\epsilon$ (3,5) Such that a = 3 and b = 5
According to MVT,

f '(c) = $= \frac{f(b) -f(a)}{b - a }$

f '(c) = $= \frac{f(5) -f(3)}{5 - 3 }$ ------ (1)

Now,
f(x) = $x^{3} -18x^{2} + 99x -162$
f(5) = $5^{3} -18(5)^{2} + 99(5) -162$
= 125 - 450 + 495 -162
f(5) = 8
f(x) = $x^{3} -18x^{2} + 99x -162$
f(3) = $3^{3} -18(3)^{2} + 99(3) -162$
= 27 - 162 + 297 -162
f(3) = 0
So equation (1) becomes
f '(c) = $= \frac{8 -0 }{5 - 3 }$
f '(c) = $= \frac{8 }{2 }$
f '(c) = 4
Now, f(x) = $x^{3} -18x^{2} + 99x -162$
f '(x) = $3x^{2} -36x + 99$
f '(c) = $3c^{2} -36c + 99$
$3c^{2} -36c + 99$ = 4
$3c^{2} -36c + 95$ = 0
Now we will use a quadratic formula to fid the value of c

$x = -b \pm\frac{\sqrt{b^{2}-4ac}}{2a}$

$c = 36 \pm\frac{\sqrt{1296-1140}}{6}$

= $36 \pm\frac{12.49}{6}$

c = 8.8 , 4.8
Thus, c = 4.8 $\epsilon$ (3,5) such that f '(c) = $= \frac{f(5) -f(3)}{5 - 3 }$
Hence MVT is verified.

Example 2: Find the value of 'c' guaranteed by the mean value theorem for f(x) = 2- $3x^{2}$ on [-1,2].
Solution : f(x) = 2- $3x^{2}$ the given function is continuous in [-1,2] and differentiable in (-1,2) so we can apply MVT to find the value of c.
f(x) = 2- $3x^{2}$
f '(x) = -6x
f '(c) = -6c
f(x) = 2- $3x^{2}$
f(-1) = 2 - $3(-1)^{2}$ = -1
f(2) = 2 - $3(2)^{2}$ = -10

f '(c) = $\frac{f(2) -f(-1)}{2 -(-1) }$

-6c = $\frac{-10 -(-1)}{2 + 1 }$

-6c = $\frac{-9}{3 }$

c = $\frac{-3}{-6 }$

c = $\frac{1}{2 }$