Question

Verify LMVT for f(x) = (x-1)(x-5)(x+10) in the interval [0,5].

Answer 1

Hint: Here we have to verify LMVT for given expression so first of all according to condition of LMVT we will check continuity in given interval and then we will check differentiability in given interval and then using formula $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ we will check There exists a c such that f’(c)= 0 where c lies in given interval.

Complete step by step answer: Lagrange’s Mean Value Theorem states that if a function f(x) is continuous on a closed interval [a,b] and differentiable on an open interval (a,b), then there is at least one point x=c in this interval such that:
                     $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
Following three conditions should satisfy for LMVT to be verified:
(i). Function f(x) is continuous on a closed interval [a,b]
(ii). Function f(x) is differentiable on an open interval (a,b).
(iii). There exists a c such that f’(c)= 0 where c lies in [a,b].
$f\left( x \right) = {x^3} + 4{x^2} - 55x + 50$
The given function f(x) is continuous in [0,5].
The given function f(c) is differentiable in [0,5]. Since, it is a polynomial function.
In the given question f(x) = (x-1)(x-5)(x+10), and a=0 and b=5.
To find f(a), we substitute the value of a which is equal to 0 in f(x),
f(a) = (0-1)(0-5)(0+10)
      = 50
To find f(b), we substitute the value of b which is equal to 5 in f(x),
f(b) = (5-1)(5-5)(5+10)
       = 0
Also, to find f’(c),
f(x) = (x-1)(x-5)(x+10)
$
   = \left( {x - 1} \right)\left( {{x^2} + 5x - 50} \right) \\
   = {x^3} + 4{x^2} - 55x + 50 \\
$
Differentiating-
f’(x) = $3{x^2} + 8x - 55$
Now, by substituting c in the above equation, we get,
f’(c) = $3{c^2} + 8c - 55$
Now, by substituting all these parameters in the formula of LMVT, we get,
$f'\left( c \right) = \dfrac{{f\left( 5 \right) - f\left( 0 \right)}}{{5 - 0}}$
3${c^2}$ + 8c -55 = $\dfrac{{0 - 50}}{5}$
3${c^2}$ + 8c -55 = -50/5
3${c^2}$ + 8c -55 = -10
3${c^2}$ + 8c -45 = 0
By solving this quadratic equation, we get,
$
  c = \dfrac{{ - 8 \pm \sqrt {{8^2} - 4 \times 3 \times ( - 45)} }}{{2 \times 8}} \\
  c = \dfrac{{ - 8 \pm \sqrt {64 + 540} }}{{16}} \\
  c = \dfrac{{ - 8 \pm \sqrt {604} }}{{16}} \\
  c = \dfrac{{ - 8 \pm 24.58}}{{16}} \\
  c = - 2.03{\text{ and }}1.03 \\
$
The interval given is [0,5] since c should lie in this interval the value of c should be 1.03 for LMVT to get verified.
So, the final value of c is 1.03

Hence, c lies between [0,5]. Therefore, LMVT is verified.

Note: We take only that value of c which satisfies LMVT by lying in the given interval. All other values are rejected. If in this case if both the values of c obtained on solving the quadratic had lied in the interval [0,5] we would have considered both.
For LMVT to satisfy at least one value of c should be obtained otherwise LMVT is not verified.