#### Question

A curve for which $$\frac{\partial y}{\partial x}=3x^{2}-\frac{2}{x^{3}}$$
passes through (-1, 3). Find the equation of the curve.

Ans:$$y=\frac{3x^3}{3}-\frac{2x^{-1}}{-2}+c$$
$$3=-1+1+c$$
$$y=x^{3}+x^{-2}+3$$

### Question

A curve has equation y = f(x), and it is given that f′ (x) $$= 2x^{2} – 7 – \frac{4}{x^{2}}$$

(a) Given that f(1) = $$-\frac{1}{3}$$, find f(x).

(b) Find the coordinates of the stationary points on the curve.

(c) Find f′′ (x).

(d) Hence, or otherwise, determine the nature of each of the stationary points.

(a)  $$f(x)=\frac{2}{3}x^{3} – 7x + 4x^{-1}[+c]$$

$$-\frac{1}{3} = \frac{2}{3} – 7 + 4 + c$$  leading to c = 

$$f(x)=\frac{2}{3}x^{3} – 7x + 4x^{-1} + 2$$

(b) 2x4 – 7x2 – 4  = 0

(2x2 + 1) (x2-4) [=0]

x = [±]

$$\left [ \frac{2}{3}\left ( 2 \right )^{3} -7(2)+\frac{4}{2}+2 leading to\right ] \left ( 2, -\frac{14}{2} \right )$$

$$\left [ \frac{2}{3}\left (-2 \right )^{3} -7(-2)+\frac{4}{-2}+2 leading to\right ] \left ( -2, \frac{26}{3} \right )$$

(c) f” (x) = 4x + 8x-3

(d) f” (2) = 9 > 0 MINIMUM at x = their 2

f”(-2) = -9< 0 MAXIMUM at x = their – 2

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 $$\frac{\mathrm{d} y}{\mathrm{d} x}$$