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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 F(X) = 2x2 − 3x + 1 on [1, 3] ?

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प्रश्न

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 theorem f(x) = 2x2 − 3x + 1 on [1, 3] ?

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उत्तर

 We have,

\[f\left( x \right) = 2 x^2 - 3x + 1\]
Since a polynomial function is everywhere continuous and differentiable.
Therefore, \[f\left( x \right)\] is continuous on \[\left[ 1, 3 \right]\] and differentiable on \[\left( 1, 3 \right)\].
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​\[c \in \left( 1, 3 \right)\] such that\[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1} = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]
Now, \[f\left( x \right) = 2 x^2 - 3x + 1\]
\[\Rightarrow f'\left( x \right) = 4x - 3\],\[f\left( 3 \right) = 10\] ,\[f\left( 1 \right) = 2 \left( 1 \right)^2 - 3\left( 1 \right) + 1 = 0\]
∴  \[f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]
\[\Rightarrow 4x - 3 = \frac{10 - 0}{2}\]

\[ \Rightarrow 4x - 3 - 5 = 0\]

\[ \Rightarrow x = 2\]

Thus, 

\[c = 2 \in \left( 1, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\]

Hence, Lagrange's theorem is verified.
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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Mean Value Theorems - Exercise 15.2 [पृष्ठ १७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 14 Mean Value Theorems
Exercise 15.2 | Q 1.05 | पृष्ठ १७

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