Question

a curve is such that dy/dx = -x square + 5x - 4.
1) find the x- coordinates of each of the stationary points of the curve.
2) obtain an expression for d square y/d x square (second derivative) and hence or otherwise find the nature of each of the stationary points .
3)given that the curve passes through the point (6 , 2 ), find the equation of the curve .

Answer 1

1) To find coordinates of stationary points we should solve $\frac{dy}{dx}=0$

$-x^2+5x-4=0$

$-(x-1)(x-4)=0$

$x=1$ or $x=4$

2) $\frac{d^2y}{dx^2}=(-x^2+5x-4)'=-2x+5$

when $x=1$ $\frac{d^2y}{dx^2}=3>0$ point of local minimum

when $x=4$ $\frac{d^2y}{dx^2}=-3<0$ point of local maximum

3) $y=\int(-x^2+5x-4)dx=\frac{-x^3}{3}+5\frac{x^2}{2}-4x+C$

$(6,2)$

$2=-\frac{6^3}{3}+5*\frac{6^2}{2}-4*6+C$

$2= -72+90-24+C$

$C=8$

$y=\frac{-x^3}{3}+5\frac{x^2}{2}-4x+8$