find the maximum and minimum values of the function for(x) =2x^2-6x+3
مهاباد (2)​

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1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Given that

   [tex]\rm :\longmapsto\:f(x) = {2x}^{2} – 6x + 3[/tex]

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

   [tex]\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}( {2x}^{2} – 6x + 3)[/tex]

   [tex]\rm :\longmapsto\:f'(x) = 2\dfrac{d}{dx} {x}^{2} – 6\dfrac{d}{dx}x + \dfrac{d}{dx}3[/tex]

   [tex]\rm :\longmapsto\:f'(x) = 4x – 6 – – – (1)[/tex]

   For maxima and minima,

   [tex]\rm :\longmapsto\:Put \: f'(x) = 0[/tex]

   [tex]\rm :\longmapsto\:4x – 6 = 0[/tex]

   [tex]\rm :\longmapsto\:4x = 6[/tex]

   [tex]\rm :\implies\:x = \dfrac{3}{2} – – – (2)[/tex]

   Now, from equation (1), we have

   [tex]\rm :\longmapsto\:f'(x) = 4x – 6[/tex]

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

   [tex]\rm :\longmapsto\:\dfrac{d}{dx}f'(x) =\dfrac{d}{dx}( 4x – 6)[/tex]

   [tex]\rm :\longmapsto\:f”(x) = 4 > 0[/tex]

   [tex]\bf\implies \:f(x) \: is \: minimum \: at \: x = \dfrac{3}{2} [/tex]

   and

   [tex]\rm :\longmapsto\:Minimum \: value \: is \: f\bigg(\dfrac{3}{2} \bigg) [/tex]

   [tex] \rm \: = \: \: 2 {\bigg(\dfrac{3}{2} \bigg) }^{2} – 6 \times \dfrac{3}{2} + 3[/tex]

   [tex] \rm \: = \: \: 2 \times \dfrac{9}{4} – 9 + 3[/tex]

   [tex] \rm \: = \: \:\dfrac{9}{2} – 6[/tex]

   [tex] \rm \: = \: \:\dfrac{9 – 12}{2}[/tex]

   [tex] \rm \: = \: \:\dfrac{ – 3}{2}[/tex]

   Basic Concept Used :-

   HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A FUNCTION

   Let given function be f(x).

   Differentiate the given function, we get f'(x)

   let f'(x) = 0 and find critical point say x = a.

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

   Apply the critical point in the second derivative.

   The function f (x) is maximum when f”(a) < 0.

   The function f (x) is minimum when f”(a) > 0.

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