Question

Local minima for f(x)= X/2 + 2/X occurs at ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{x}{2} + \dfrac{2}{x}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) =\dfrac{d}{dx}\bigg[ \dfrac{x}{2} + \dfrac{2}{x}\bigg]$

We know,

$\boxed{\tt{\dfrac{d}{dx}{x}^{n} = {nx}^{n - 1}}}$

and

$\boxed{\tt{ \dfrac{d}{dx} \frac{1}{ {x}^{n} } = - \frac{n}{ {x}^{n + 1} }}}$

So, using this, we get

$\rm :\longmapsto\:f'(x) = \dfrac{1}{2} - \dfrac{2}{ {x}^{2} }$

For, maxima or minima, we get

$\rm :\longmapsto\:f'(x) = 0$

$\rm :\longmapsto\:\dfrac{1}{2} - \dfrac{2}{ {x}^{2} } = 0$

$\rm :\longmapsto\: \dfrac{ {x}^{2} - 4}{ 2{x}^{2} } = 0$

$\rm :\longmapsto\: {x}^{2} - 4 = 0$

$\rm :\longmapsto\:(x - 2)(x + 2) = 0$

$\bf\implies \:x = 2 \: \: or \: \: x = - 2$

Now, We have

$\rm :\longmapsto\:f'(x) = \dfrac{1}{2} - \dfrac{2}{ {x}^{2} }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f'(x) =\dfrac{d}{dx}\bigg( \dfrac{1}{2} - \dfrac{2}{ {x}^{2} } \bigg)$

$\rm :\longmapsto\:f''(x) = 0 + \dfrac{2}{ {x}^{3} }$

$\rm :\longmapsto\:f''(x) = \dfrac{2}{ {x}^{3} }$

When x = 2

$\rm :\longmapsto\:f''(2) = \dfrac{2}{ {2}^{3} } = \dfrac{1}{4} > 0$

$\rm\implies \:f(x) \: is \: minimum \: at \: x = 2$

Now, When x = - 2

$\rm :\longmapsto\:f''( - 2) = \dfrac{2}{ {( - 2)}^{3} } = \dfrac{1}{ - 4} < 0$

$\rm\implies \:f(x) \: is \: maximum \: at \: x = - 2$

Hence, f(x) is minimum at x = 2

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Basic Concept Used :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

• The function f (x) is maximum when f''(x) < 0.

• The function f (x) is minimum when f''(x) > 0.

Answer 2

Question :-

• Local minima for f(x)= $\frac{x}{2}+ \frac{2}{x}$occurs at ?

Solution :-

$\: \: \: \: : \rightarrow \mathtt{f(x) = \frac{x}{2} + \frac{2}{x} } \\ \: \: \: \: \: \: \: : \rightarrow \mathtt{f {}^{′}(x) = \frac{1}{2} + 2x - \frac{1}{ {x}^{2} } } \\ \: \: \: \: \: \: \: \: \: : \longmapsto \mathtt{ {f}^{′} (x) = \frac{1}{2} - \frac{2}{x {}^{2} } } \\ \implies \mathtt{ \frac{ {x}^{2} - 4 }{2 {x}^{2} } }$

Now f ' (x) = 0 which is implies

$\: \: \: \: \: \: : \implies \mathtt{ \frac{ {x}^{2} - 4}{ {2x}^{2} } = 0 } \\ : \rightarrow \mathtt{ {x}^{2} - 4} \\ \: \: \: \hookrightarrow \boxed{\mathtt{x =± \: 2 }}$

So, at x = 2

$\: \: \: \: \: \: \: \rightarrow \: f \mathtt{(2) = \frac{2}{2} + \frac{2}{2} } \\ \: \: \: \: \: \: \: \rightarrow \: f \mathtt{(2) = 1 + 1} \\ \: \: \: \: \: \: \hookrightarrow \boxed{ f\mathtt{(2) = 2}}$

$\red{\bigstar}$ Local minima is at x = 2 and f(2) = 2 .