Question

The value of c in Lagrange's mean value theorem for the function f (x) = x (x − 2) when x ∈ [1, 2] is

Options

• 1

• 1/2

• 2/3

• 3/2

Answer 1

\[\frac{3}{2}\]

We have
 f (x) = x (x − 2)

It can be rewritten as \[f\left( x \right) = x^2 - 2x\] .

We know that a polynomial function is everywhere continuous and differentiable.
Since \[f\left( x \right)\] is a polynomial , it is continuous on \[\left[ 1, 2 \right]\] and differentiable on \[\left( 1, 2 \right)\] .

Thus, \[f\left( x \right)\] satisfies both the conditions of Lagrange's theorem on \[\left[ 1, 2 \right]\] .

So, there must exist at least one real number c \[\left[ 1, 2 \right]\] such that 

\[f'\left( c \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1} = \frac{f\left( 2 \right) - f\left( 1 \right)}{1}\]

Now, \[f\left( x \right) = x^2 - 2x\] 

\[\Rightarrow f'\left( x \right) = 2x - 2\], and\[f\left( 1 \right) = - 1, f\left( 2 \right) = 0\]

\[\therefore f'\left( x \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}\]

\[\Rightarrow f'\left( x \right) = \frac{0 + 1}{1}\]

\[ \Rightarrow 2x - 2 = 1\]

\[ \Rightarrow x = \frac{3}{2}\]

∴ \[c = \frac{3}{2} \in \left( 1, 2 \right)\].