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F(X) = X3 (X − 1)2 . - Mathematics

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Question

f(x) = x3  (x \[-\] 1).

Solution

\[\text { Given }: f\left( x \right) = x^3 \left( x - 1 \right)^2 \]

\[ \Rightarrow f'\left( x \right) = 3 x^2 \left( x - 1 \right)^2 + 2 x^3 \left( x - 1 \right)\]

\[\text { For a local maximum or a local minimum, we must have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow 3 x^2 \left( x - 1 \right)^2 + 2 x^3 \left( x - 1 \right) = 0\]

\[ \Rightarrow x^2 \left( x - 1 \right)\left\{ 3x - 3 + 2x \right\} = 0\]

\[ \Rightarrow x^2 \left( x - 1 \right)\left( 5x - 3 \right) = 0\]

\[ \Rightarrow x = 0, 1, \frac{3}{5}\]

Since f '(x) changes from negative to positive when x increases through 1, x = 1 is the point of local minima.

The local minimum value of  f (x)  at x = 1 is given by \[\left( 1 \right)^3 \left( 1 - 1 \right)^2 = 0\]

Since f '(x) changes from positive to negative when x increases through \[\frac{3}{5}\], x = \[\frac{3}{5}\] is the point of local maxima.

The local minimum value of  f (x) at x =  \[\frac{3}{5}\] is given by \[\left( \frac{3}{5} \right)^3 \left( \frac{3}{5} - 1 \right)^2 = \frac{27}{125} \times \frac{4}{25} = \frac{108}{3125}\]
Sincef '(x) does not change from positive as x increases through 0, x = 0 is a point of inflexion.

Notes

The solution in the book is incorrect. The solution here is created according to the question given in the book.

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APPEARS IN

 RD Sharma Solution for Mathematics for Class 12 (Set of 2 Volume) (2018 (Latest))
Chapter 18: Maxima and Minima
Ex. 18.2 | Q: 3 | Page no. 16
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F(X) = X3 (X − 1)2 . Concept: Graph of Maxima and Minima.
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