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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) = X2 + X − 1 on [0, 4] ?

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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) = x2 + x − 1 on [0, 4] ?

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उत्तर

We have,

\[f\left( x \right) = x^2 + x - 1\]

Since 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 the conditions of lagrange's theorem are satisfied.
Consequently, there exists some \[c \in \left( 0, 4 \right)\] 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, \[f\left( x \right) = x^2 + x - 1\]
\[f'\left( x \right) = 2x + 1\],\[f\left( 4 \right) = 19\],\[f\left( 0 \right) = - 1\]
∴ \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\]

\[\Rightarrow 2x + 1 = \frac{20}{4}\]

\[ \Rightarrow 2x + 1 = 5\]

\[ \Rightarrow 2x = 4 \]

\[ \Rightarrow x = 2\]

Thus, \[c = 2 \in \left( 0, 4 \right)\] such that \[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\].

Hence, Lagrange's theorem is verified.
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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 1.14 | पृष्ठ १७

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