PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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Solution for Verify Lagrange'S Mean Value Theorem for the Following Function on the Indicated Intervals. Find a Point 'C' in the Indicated Interval as Stated by the F(X) = (X − 1)(X − 2)(X − 3) on [0, 4] ? - PUC Karnataka Science Class 12 - Mathematics

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Question

Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theore f(x) = (x − 1)(x − 2)(x − 3) on [0, 4] ?

Solution

 We have,

\[f\left( x \right) = \left( x - 1 \right)\left( x - 2 \right)\left( x - 3 \right)\] which can be rewritten as

\[f\left( x \right) = x^3 - 6 x^2 + 11x - 6\]
Since a polynomial function is everywhere continuous and differentiable.
Therefore, 
\[f\left( x \right)\] is continuous on \[\left[ 0, 4 \right]\] and differentiable on \[\left( 0, 4 \right)\] .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ \[c \in \left( 0, 4 \right)\] such that
\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0} = \frac{f\left( 4 \right) - f\left( 0 \right)}{4}\]
Now, 
\[f\left( x \right) = x^3 - 6 x^2 + 11x - 6\]
\[\Rightarrow f'\left( x \right) = 3 x^2 - 12x + 11\] ,
\[f\left( 0 \right) = - 6\] ,
\[f\left( 4 \right) = 64 - 96 + 44 - 6 = 6\]
∴ \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\]

\[\Rightarrow 3 x^2 - 12x + 11 = \frac{6 + 6}{4}\]

\[ \Rightarrow 3 x^2 - 12x + 8 = 0\]

\[ \Rightarrow x = 2 - \frac{2}{\sqrt{3}}, 2 + \frac{2}{\sqrt{3}}\]

Thus, 

\[c = 2 \pm \frac{2}{\sqrt{3}} \in \left( 0, 4 \right)\] such that

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\] .

Hence, Lagrange's theorem is verified.

  Is there an error in this question or solution?
Solution Verify Lagrange'S Mean Value Theorem for the Following Function on the Indicated Intervals. Find a Point 'C' in the Indicated Interval as Stated by the F(X) = (X − 1)(X − 2)(X − 3) on [0, 4] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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