Question

It is given that a curve has equation $$y=f\left ( x \right )$$  ,where $$f\left ( x \right )=x^{3}-2x^{2}+x$$

(i)Find the set of values of x for which the gradient of the curve is less than 5.

(ii)Find the values of f(x) at the two stationary points on the curve and determine the nature if each stationary point.

(i)$$3x^{2}-4x+1< 5$$

$$\left ( 3x+2 \right )\left ( x-2 \right )< 0$$

$$-\frac{2}{3}< x< 2$$ or $$\left [ -\frac{2}{3} ,2\right ]$$ or $$\left ( -\frac{2}{3} ,2\right )$$

(ii)$$3x^{2}-4x+1=0\Rightarrow \left ( 3x-1 \right )\left ( x-1 \right )=0$$

$$x=\frac{1}{3}$$ or 1

$$y=\frac{4}{27} or 0 \(f{}”\left ( x \right )=6x-4\rightarrow f{\left ( \frac{1}{3} \right )}”=-2< 0$$

$$f{}”\left ( 1 \right )=2> 0$$

max at $$\left ( \frac{1}{3},\frac{4}{27} \right )$$ ;min at (1,0)

Question

A curve for which $$\frac{\partial^2 y}{\partial x^2} = 2x − 5$$ has a stationary point at (3 , 6 )

(i) Find the equation of the curve.

(ii) Find the x-coordinate of the other stationary point on the curve.

(iii) Determine the nature of each of the stationary points.

(i) integrating → $$\frac{dy}{dx}=x^{2}− 5x (+c)$$

= 0 when x = 3

c = 6

integrating again → $$y=\frac{x^{3}}{3}-\frac{5x^{2}}{2}+6x+d$$

use of (3, 6)  d=1.5

(ii)  $$\frac{\mathrm{d} y}{\mathrm{d} x}=x^{2}-5x+6\rightarrow x=2$$

(iii) $$x=3, \frac{\partial^2 y}{\partial x^2}=1$$ and/or +ve Minimum

$$x=2, \frac{\partial^2 y}{\partial x^2}=-1$$ and /or -ve Maximum

May use shape of$$+x^3$$ curve or change in sign of $$\frac{\mathrm{d} y}{\mathrm{d} x}$$