The number of critical point of f ( x ) = x 2 ∣ x − 1∣ is

f ( x ) = ± ( x − x 1 ) ⇒ f ′ ( x ) = ± ( 1 + x 2 1 )  = 0

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MathematicsApplication of DerivativeMaxima-MinimaMedium2 minQB

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The number of critical point of $f (x) = \frac{∣ x - 1∣}{x ^{2}}$ is

Answer:

Solution:

$f (x) = \pm (x - \frac{1}{x}) \Rightarrow f^{'} (x) = \pm (1 + \frac{1}{x ^{2}}) \neq = 0$

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

⏱ 2mℹ️ Source: QB

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