Math, asked by sakshikadian71, 3 months ago

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

Answered by mathdude500
4

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

Given that

\rm :\longmapsto\:f(x) =  {2x}^{2} - 6x + 3

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

\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}( {2x}^{2} - 6x + 3)

\rm :\longmapsto\:f'(x) = 2\dfrac{d}{dx} {x}^{2} - 6\dfrac{d}{dx}x + \dfrac{d}{dx}3

\rm :\longmapsto\:f'(x) = 4x - 6 -  -  - (1)

For maxima and minima,

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

\rm :\longmapsto\:4x - 6 = 0

\rm :\longmapsto\:4x = 6

\rm :\implies\:x = \dfrac{3}{2}  -  -  - (2)

Now, from equation (1), we have

\rm :\longmapsto\:f'(x) = 4x - 6

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

\rm :\longmapsto\:\dfrac{d}{dx}f'(x) =\dfrac{d}{dx}( 4x - 6)

\rm :\longmapsto\:f''(x) = 4 > 0

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

and

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

 \rm \:  =  \:  \: 2 {\bigg(\dfrac{3}{2}  \bigg) }^{2}  - 6 \times \dfrac{3}{2}  + 3

 \rm \:  =  \:  \: 2 \times \dfrac{9}{4}  - 9 + 3

 \rm \:  =  \:  \:\dfrac{9}{2}  - 6

 \rm \:  =  \:  \:\dfrac{9 - 12}{2}

 \rm \:  =  \:  \:\dfrac{ - 3}{2}

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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