(B) Given $f(x) = x^2 + \frac{54}{x}$.First,find the derivative: $f'(x) = 2x - \frac{54}{x^2}$.For $x \in (-\infty, 0)$,$x^2 > 0$ and $x < 0$,so $2x <

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is decreasing and has neither maximum nor minimum in the interval $(-\infty, 0)$

(B) Given $f(x) = x^2 + \frac{54}{x}$.First,find the derivative: $f'(x) = 2x - \frac{54}{x^2}$.For $x \in (-\infty, 0)$,$x^2 > 0$ and $x For critical points,set $f'(x) = 0$: $2x - \frac{54}{x^2} = 0 \Rightarrow 2x^3 = 54 \Rightarrow x^3 = 27 \Rightarrow x = 3$.Since the only critical point $x = 3$ lies in $(0, \infty)$,there is no critical point in $(-\infty, 0)$.Therefore,$f(x)$ has neither maximum nor minimum in $(-\infty, 0)$.

Class 12 Mathematics Applications of Derivatives Maxima and Minima

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The function $f(x) = x^2 + \frac{54}{x}$

AP EAMCET 2023 All AP EAMCET

A
is increasing and has minimum value $27$ in the interval $(0, \infty)$
B
is decreasing and has neither maximum nor minimum in the interval $(-\infty, 0)$
C
has maximum value $27$ in the interval $(-\infty, \infty)$
D
is increasing and has neither maximum nor minimum values in the interval $(-\infty, \infty)$

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Applications of Derivatives

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Maxima and Minima

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The function $f(x) = x^2 + \frac{54}{x}$

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