Question

4 A curve has equation y=x^3 - 2x² +5. Find the coordinates of its two stationary points.​

Answer 1

Answer:

Stationary points are (0, 5) and (4/3, 103/27)

Step-by-step explanation:

At stationary point (turning point), dy/dx :

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

but dy/dx = 0:

$x(3x - 4) = 0 \\ either \: x \: is \: 0 \: and \: (3x - 4) = 0 \\ x = 0 \\ and \\ 3x - 4 = 0 \\ x = \frac{4}{3}$

let y = x³ - 2x² + 5:

$when \: x \: is \: 0 \\ y = 5 \\ when \: x \: is \: \frac{4}{3} \\ \\ y = {( \frac{4}{3}) }^{3} - 2 {( \frac{4}{3} )}^{2} + 5 \\ \\ y = \frac{103}{27}$