Find the maximum and minimum values of the curve y=2x³-3x²-12x+10

Mathematics

1 answer:

Ray Of Light [21]2 years ago

5 0

$\underline{ \orange{\huge \boxed{ \frak{Answer : }}}}$

Let ,

$\sf \large \color{purple} y = 2 {x}^{3} - 3 {x}^{2} - 12x + 10 \: --( \: 1 \: )$

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Now , Diff wrt ' x ' , we get :

$\sf \: \frac{dy}{dx} = \frac{d}{dx} (2 {x}^{3} - 3 {x}^{2} - 12x + 10) \\ \sf \: \sf \: \frac{dy}{dx} = \frac{d}{dx} \: 2(3 {x}^{2} ) - \frac{d}{dx} 3 {x}^{2} - \frac{d}{dx} 12x + \frac{d}{dx} 10 \\ \sf \: \frac{dy}{dx} =2(3 {x}^{2} ) - 3(2x) - 12(1) + 0 \\ \sf \: \frac{dy}{dx} =6 {x}^{2} - 6x - 12 + 0 \\ \: \sf \red{\frac{dy}{dx} = 6 {x}^{2} 6x - 12 -- (2)}$

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For maxima or minima \frac{dy}{dx} = 0

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$\sf \: 6 {x}^{2} - 6x - 12 = 0$

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Divided by 6 on both side , we get.

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$\sf \: {x}^{2} - x - 2 = 0 \\ \sf \: {x}^{2} - 2x + x - 2 = 0 \\ \sf \: x(x - 2) + 1(x - 2) = 0 \\ \sf \: (x - 2)(x + 1) = 0 \\ \sf \: x - 2 = 0 \: \: \bold or \: \: x + 1 = 0 \\ \sf \fbox{x = 2 \: } \: \bold or \: \fbox{ x = - 1}$

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Again Diff wrt ‘ x ’ , we get.

$\sf \: \frac{d}{dx} =(\frac{dy}{dx} ) = 6\frac{d}{dx} - 6\frac{d}{dx}x - \frac{d}{dx}12 \\ \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = 6(2x) - 6(1) - 0 \\ \sf \: \sf \bold{ \frac{ {d}^{2}y }{ {dx}^{2} } =12x - 6}$

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At x = 2

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$\sf \: \frac{ {d}^{2}y }{ {dx}^{2} } =12(2) - 6 \\ \: \: \: \sf \: = 24 - 6 \\ \: \: \: \: \sf \red{ = 18 > 0}$

At x = -1

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$\sf \: \frac{ {d}^{2}y }{ {dx}^{2} } =12( - 1) - 6 \\ \: \: \: \sf \: = - 12 - 6 \\ \: \: \: \: \sf \red{ = - 18 < 0 }$

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x = 2 gives minima value of function.

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x = -1 gives maxima value of function.

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Now, put x = 2 in eqⁿ ( 1 )

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$\sf \: y \: minima \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( {2})^{3} - 3 ({2})^{2} - 12(2) + 10 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: = 2(8) - 3(4) - 24 + 10 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: = 16 - 12 - 24 + 10 \\\sf \: \: \: \: \: \: \: \: \: \: = - 20 + 10 \\\sf \color{red}{\boxed{ = - 10}}$

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The Point of minima is ( 2 , -10 ).

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Now , put x = -1 in eqⁿ ( 1 )

$\sf \: y \: maxima \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( { - 1})^{3} - 3 ({ - 1})^{2} - 12( - 1) + 10 \\\sf \color{red}{\boxed{ = 17}}$