Question

find maxima and minima to given curve y=x ³-x²-x-1​

Answer 1

$\huge{\mathbb{\red{ANSWER:-}}}$

$\sf{y = x^{3} - x^{2} - x - 1}$

Differential with respect to x

$\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}(x^{3}-x^{2}-x-1)}$

$\sf{\dfrac{dy}{dx}= 3x^{2} - 2x - 1}$

$\sf{For \: roots-}$

$\sf{\dfrac{dy}{dx} = 0}$

$\sf{3x^{2} - 2x - 1 = 0}$

$\sf{3x^{2} - 3x + x - 1}$

$\sf{3x(x - 1) + 1(x - 1) = 0}$

$\sf{(x - 1)(3x + 1) = 0}$

$\sf{x = 1 \: , \: x = \dfrac{-1}{3}}$

$\sf{Now \: ,}$

$\sf{\dfrac{d^{2} y}{dx^{2}} =\dfrac{d}{dx}(\dfrac{dy}{dx})}$

$\sf{\dfrac{d^{2} y}{dx^{2}} =\dfrac{d}{dx}(3x^{2} - 2x - 1)}$

$\sf{\dfrac{d^{2} y}{dx^{2}} = 6x - 2}$

$\sf{For \: any \: value \: of \: x-}$

$\sf{if \: \dfrac{d^{2} y}{dx^{2}} < 0 \: ,}$

$\sf{then \: , \: there \: would \: be \: maxima.}$

$\sf{if \: \dfrac{d^{2} y}{dx^{2}} > 0 \: ,}$

$\sf{then \: , \: there \: would \: be \: minima.}$

$\sf{At \: x = 1}$

$\sf{\dfrac{d^{2} y}{dx^{2}} = 6 - 2 = 4 > 0}$

$\sf{At \: x =\dfrac{-1}{3}}$

$\sf{\dfrac{d^{2} y}{dx^{2}} =6(\dfrac{-1}{3}) - 2 = -2-2 = -4 < 0}$

$\sf{So \: ,}$

$\sf{At \: x = 1 \: , \: There \: is \: minima.}$

$\sf{At \: x =\dfrac{-1}{3} \: , \: There \: is \: maxima.}$