Question

Pls help...pls pls pls pls pls pls solve these

Answer 1

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Answer 2

$\text{At maxima,}\\ \frac{dy}{dx} = 0 \: \: \: and \: \: \: \frac{ {d}^{2} y}{d {x}^{2} } = - ve$

$\text{At minima,}\\ \frac{dy}{dx} = 0 \: \: \: and \: \: \: \frac{ {d}^{2} y}{d {x}^{2} } = + ve$

$i) \: y = - {(9x + 2)}^{2} \\ y = - (81 {x}^{2} + 36x + 4) \\ y = - 81 {x}^{2} - 36x - 4$
$\frac{dy}{dx} = - 81 \times 2x - 36 \times 1 = 0 \\ - 162x - 36 = 0 \\ - 162x = 36 \\ x = - \frac{36}{162} = - \frac{2}{9}$
$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{d}{dx} ( - 162x - 36) = - 162 \\ \\ since \: \frac{ {d}^{2} y}{d {x}^{2} } = - ve \\ it \: is \: maxima.$
$y = - {(9x + 2)}^{2} = - (9 \times \frac{ - 2}{9} + 2)^{2} = 0$
$\text{Thus position of maxima is} \: ( \frac{ - 2}{9} . \: 0)$

Similarly you can do the rest.