Question

find any maxima, minima or horizontal points of inflexion of the curve y=(x^3+3x-1)/x^2 stating with reasons, the nature of each point.

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{ {x}^{3} + 3x - 1 }{ {x}^{2} }$

$\rm :\longmapsto\:f(x) = x + \dfrac{3}{x} - \dfrac{1}{ {x}^{2} }$

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

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) =\dfrac{d}{dx}\bigg[x + \dfrac{3}{x} - \dfrac{1}{ {x}^{2} }\bigg]$

$\rm :\longmapsto\:f'(x) = \dfrac{d}{dx}x + 3\dfrac{d}{dx}\dfrac{1}{x} - \dfrac{d}{dx} \dfrac{1}{ {x}^{2} }$

We know,

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

So, using this, we get

$\rm :\longmapsto\:f'(x) = 1 - \dfrac{3}{ {x}^{2} } + \dfrac{2}{ {x}^{3} }$

For maxima or minima, Substitute

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

$\rm :\longmapsto\:1 - \dfrac{3}{ {x}^{2} } + \dfrac{2}{ {x}^{3} } = 0$

$\rm :\longmapsto\:\dfrac{ {x}^{3} - 3x + 2 }{ {x}^{3} } = 0$

$\rm :\longmapsto\: {x}^{3} - 3x + 2 = 0$

$\rm :\longmapsto\: {x}^{3} - x - 2x + 2 = 0$

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

$\rm :\longmapsto\:x(x - 1)(x + 1) - 2(x - 1) = 0$

$\rm :\longmapsto\:(x - 1)( {x}^{2} + x - 2) = 0$

$\rm :\longmapsto\:(x - 1)( {x}^{2} + 2x - x - 2) = 0$

$\rm :\longmapsto\:(x - 1)[x(x + 2) - 1(x + 2)] = 0$

$\rm :\longmapsto\:(x - 1)(x - 1)(x + 2) = 0$

$\bf\implies \:x = 1 \: \: or \: \: x \: = - \: 2$

Now, we have

$\rm :\longmapsto\:f'(x) = 1 - \dfrac{3}{ {x}^{2} } + \dfrac{2}{ {x}^{3} }$

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

$\rm :\longmapsto\:\dfrac{d}{dx}f'(x) =\dfrac{d}{dx}\bigg[ 1 - \dfrac{3}{ {x}^{2} } + \dfrac{2}{ {x}^{3} }\bigg]$

$\rm :\longmapsto\:f''(x) = \dfrac{6}{ {x}^{3} } - \dfrac{6}{ {x}^{4} }$

$\rm :\longmapsto\:f''(x) = \dfrac{6(x - 1)}{ {x}^{4} }$

Consider, When x = 1

$\rm :\longmapsto\:f''(1) = \dfrac{6(1 - 1)}{ {1}^{4} } = 0$

$\rm\implies \:x = 1 \: is \: the \: point \: of \: inflexion$

Consider, When x = - 2

$\rm :\longmapsto\:f''(2) = \dfrac{6( - 2 - 1)}{ {( - 2)}^{4} } = - 18 < 0$

$\rm\implies \:x = - 2\: is \: the \: point \: of \: maxima$

Thus we have

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = 1 \: is \: the \: point \: of \: inflexion} \\ \\ &\sf{x = - 2 \: is \: point \: of \: maxima} \end{cases}\end{gathered}\end{gathered}}$

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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 f''(x), to check the nature of points.

5. Condition :-

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

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

• The function f(x) have point of inflexion when f''(x) = 0