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in Derivatives by (28.3k points)
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The least value of the function f(x) = \(\frac{x}{4-x+x^2}\) on [–1, 1] is :

A. \(-\frac{1}{4}\)

B. \(-\frac{1}{3}\)

C. \(-\frac{1}{6}\)

D. \(\frac{1}{5}\)

1 Answer

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Best answer

Option : (C)

⇒ 4 - x= 0

= x = ± 2 but [-1,1]

= f(-1) =\(-\frac{1}{6}\)

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