The set of critical points of the function f ( x ) = ( x − 2 ) 3 2 ⋅ ( 2 x + 1 ) is

f ′ ( x ) = ( x − 2 ) 3 2 ⋅ 2 + ( 2 x + 1 ) ⋅ 3 2 ( x − 2 ) 3 1 1 = 2 [ 3 ( x − 2 ) 3 1 3 ( x − 2 ) + 2 x + 1 ] = 3 2 ( x − 2 ) 3 1 ( 5 x − 5 ) = 3 10 ( x − 2 ) 3 1 ( x − 1 ) Critical points are X = 1 and X = 2

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaEasy2 minQB

MathematicsEasysingle choice

The set of critical points of the function $f (x) = (x - 2)^{\frac{2}{3}} \cdot (2 x + 1)$ is

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Solution:

$f^{'} (x) = (x - 2)^{\frac{2}{3}} \cdot 2 + (2 x + 1) \cdot \frac{2}{3} \frac{1}{( x - 2 ) ^{\frac{1}{3}}}$ $= 2 [\frac{3 ( x - 2 ) + 2 x + 1}{3 ( x - 2 ) ^{\frac{1}{3}}}] = \frac{2}{3} \frac{( 5 x - 5 )}{( x - 2 ) ^{\frac{1}{3}}} = \frac{10}{3} \frac{( x - 1 )}{( x - 2 ) ^{\frac{1}{3}}}$

Critical points are $X = 1$ and $X = 2$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

⏱ 2mℹ️ Source: QB

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