Question

– If y = f(x) is continuous on [0, 6]. Differentiable on (0, 6). If f (0) = -2 and f (6) = 16, then at some point between x = 0 and x = 6, f’(x) must be equal to
$\left( A \right)$ -18
$\left( B \right)$ -3
$\left( C \right)$ 3
$\left( D \right)$ 14
$\left( E \right)$ 18

Answer 1

Hint – In this particular type of question use the concept that if a function is continuous in the given interval say [a, b] then there exists a point c between the interval such that $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ according to Lagrange’s mean value theorem so use this concept to reach the solution of the question.

Complete step-by-step answer:
Given data:
Y = f (x) is a continuous on [0, 6]. Differentiable on (0, 6).
And f (0) = -2 and f (6) = 16.................. (1)
Then we have to find out the value of f’(x) at some point between (0 and 6)
So f is continuous so according to Lagrange’s mean value theorem there exists a point c in between the limits of f such that,
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ Where, b and a are the final and initial limits of f.
Now substitute the values we have,
 \[ \Rightarrow f'\left( c \right) = \dfrac{{f\left( 6 \right) - f\left( 0 \right)}}{{6 - 0}}\]
Now from equation (1) we have,
$ \Rightarrow f'\left( c \right) = \dfrac{{16 - \left( { - 2} \right)}}{6} = \dfrac{{18}}{6} = 3$
So in place of c substitute x we have,
$ \Rightarrow f'\left( x \right) = 3$
So the value of f’(x) = 3 at some point between x = 0 and x = 6.
So this is the required answer.
Hence option (C) is the correct answer.

Note – Whenever we face such types of questions always remember the condition of continuous function which is stated above (don’t confuse between the range i.e. interval in which the function is continuous and the initial and final limits of the function both are same) then simply substitute all the values in the Lagrange’s mean value theorem equation and simplify as above we will get the required answer.