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The value of c in Rolle's theorem for the function f (x) = x3 − 3x in the interval [0, √ 3 ] is

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प्रश्न

The value of c in Rolle's theorem for the function f (x) = x3 − 3x in the interval [0,\[\sqrt{3}\]] is 

 

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उत्तर

 1
The given function is \[f\left( x \right) = x^3 - 3x\] .

This is a polynomial function, which is continuous and derivable in R.
Therefore, the function is continuous on [0,\[\sqrt{3}\]] and derivable on (0,\[\sqrt{3}\])

Differentiating the given function with respect to x, we get 

\[f'\left( x \right) = 3 x^2 - 3\]

\[ \Rightarrow f'\left( c \right) = 3 c^2 - 3\]

\[ \therefore f'\left( c \right) = 0 \]

\[ \Rightarrow 3 c^2 - 3 = 0\]

\[ \Rightarrow c^2 = 1\]

\[ \Rightarrow c = \pm 1\]

Thus, \[c = 1 \in \left[ 0, \sqrt{3} \right]\] for which Rolle's theorem holds.

Hence, the required value of c is 1.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Mean Value Theorems - Exercise 15.4 [पृष्ठ २०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 14 Mean Value Theorems
Exercise 15.4 | Q 10 | पृष्ठ २०

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