The minimum value of the function fx- x3-3 -x occurs at -

F(x)=x^33 x then dfdx=x^2 1,d^2fdx^2=2x For critical POint dfdx=0⇒ x=± 1 At x=1,d^2fdx^2=2(+ve) i.e., Minima occurs At x= 1,d^2fdx^2= 2( ve) i.e., Maxima occurs ...

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Standard XII

Mathematics

Integration by Substitution

The minimum v...

Question

The minimum value of the function $f (x) = (\frac{x^{3}}{3}) - x$ occurs at :

x = 1

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x = - 1

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x = 0

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x = \frac{1}{\sqrt{3}}

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Solution

The correct option is A $x = 1$
$f (x) = \frac{x^{3}}{3} - x$ then $\frac{d f}{d x} = x^{2} - 1, \frac{d^{2} f}{d x^{2}} = 2 x$
For critical POint $\frac{d f}{d x} = 0 \Rightarrow x = \pm 1$
At $x = 1, \frac{d^{2} f}{d x^{2}} = 2 (+ v e) i . e .,$ Minima occurs
At $x = - 1, \frac{d^{2} f}{d x^{2}} = - 2 (- v e) i . e .,$ Maxima occurs.
So, $x = 1$ is Point of Minima.

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The minimum value of the function f(x)=(x33)−x occurs at :

The correct option is A x=1f(x)=x33−x then dfdx=x2−1,d2fdx2=2x For critical POint dfdx=0⇒x=±1 At x=1,d2fdx2=2(+ve)i.e., Minima occurs At x=−1,d2fdx2=−2(−ve)i.e., Maxima occurs. So, x=1 is Point of Minima.

The minimum value of the function $f (x) = (\frac{x^{3}}{3}) - x$ occurs at :