Define f ( 1 ) = 8 and f ( x ) = ∣ x − 1∣ + 5 for x  = 1 . At x = 1 , the function has

For all sufficiently close x  = 1 , f ( x ) = 5 + ∣ x − 1∣ < 8 = f ( 1 ) . Hence x = 1 is a strict local maximum.

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaEasy2 minQB

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Define $f (1) = 8$ and $f (x) = ∣ x - 1∣ + 5$ for $x \neq = 1$ . At $x = 1$ , the function has

Answer:

Solution:

For all sufficiently close $x \neq = 1, f (x) = 5 + ∣ x - 1∣ < 8 = f (1)$ . Hence $x = 1$ is a strict local maximum.

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

⏱ 2mℹ️ Source: QB

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