# Mean Value Theorem

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

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 = $\frac{1}{2 }$