Question

Find the value of c if LMVT conditions are satisfied for $2{x^2} - 7x + 10,{\text{ }}x \in \left[ {2,5} \right]$.

Answer 1

Hint: Use the concept that $2{x^2} - 7x + 10$ is both differentiable as well as continuous in the interval [2, 5], so according to Lagrange’s mean value theorem there exists a point c such that $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$, where b=5 and a=2.

Complete step-by-step answer:

Lagrange’s mean value theorem (LMVT) states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
Now given function is
$f\left( x \right) = 2{x^2} - 7x + 10,{\text{ }}x \in \left[ {2,5} \right]$
Now as we know that f(x) is differentiable as well as continuous in the interval [2, 5] so there exists a point x = c such that
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ ....................... (1) Where, (a = 2, b =5)
Now differentiate f(x) we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {2{x^2} - 7x + 10} \right) = 4x - 7 + 0$
$ \Rightarrow f'\left( x \right) = 4x - 7$
Now in place of x substitute (c) we have,
$ \Rightarrow f'\left( c \right) = 4c - 7$
Now from equation (1) we have,
$ \Rightarrow 4c - 7 = \dfrac{{\left( {2{{\left( 5 \right)}^2} - 7\left( 5 \right) + 10} \right) - \left( {2{{\left( 2 \right)}^2} - 7\left( 2 \right) + 10} \right)}}{{5 - 2}}$
Now simplify the above equation we have,
$ \Rightarrow 4c - 7 = \dfrac{{\left( {50 - 35 + 10} \right) - \left( {8 - 14 + 10} \right)}}{3} = \dfrac{{25 - 4}}{3} = \dfrac{{21}}{3} = 7$
$ \Rightarrow 4c = 7 + 7$
$ \Rightarrow c = \dfrac{{14}}{4} = \dfrac{7}{2}$
Hence the value of c is (7/2).
So this is the required answer.

Note: If a function is continuous at some points then it may or may not be differentiable at those points, but if a function is differentiable at some points that we can say with certainty that it has to be continuous. That is differentiability is a sure condition for continuity however converse is not true. These tricks help commenting upon continuity and differentiability while solving problems of such kind.