Question

A curve has equation $$y=f\left ( x \right )$$  and is such that  $$f{}’\left ( x \right )=3x^{\frac{1}{2}}+3x^{-\frac{1}{2}}-10$$.

(i)By using the substitution ,or ot6herwise,find the value of x for which the curve y=f(x has stationary points.

(ii) Find and hence ,or otherwise ,determine the nature of each stationary point.

(iii)It is given that the curve y=f(x) passes through the point (4,-7) .Find f(x).

(i)$$3u+\frac{3}{u}-10=0$$

$$3u^{2}-10u+3=0\Rightarrow \left ( 3u-1 \right )\left ( u-3 \right )=0$$

$$\sqrt{x}=\frac{1}{3}$$ or  3

$$\sqrt{x}=\frac{1}{9}or 9$$

(ii)$$f{}”\left ( x \right )=\frac{3}{2}x^{-\frac{1}{2}} -\frac{3}{2} x^{-\frac{3}{2}}$$

At $$x=\frac{1}{9}$$

$$f{}”\left ( x \right )=\frac{3}{2}\left ( 3 \right )-\frac{3}{2}\left ( 27 \right )=-36< 0\rightarrow$$Max

At x=9

$$f{}”\left ( x \right )=\frac{3}{2}\times \frac{1}{3}-\frac{3}{2}\times \frac{1}{27}=\frac{4}{9}> 0\rightarrow$$Min

(iii)$$f(x)=2x^{\frac{3}{2}}+6x^{\frac{1}{2}}-10x(+c)$$

-7=16+12-40+c

c=5

Question

The function f is such that $$f{}'(x)=3x^{2}-7$$ and f . Find $$f\left ( x \right )$$.

$$f(x)=x^{3}-7x+c$$

5=27-21+c

c=-1$$\rightarrow f(x)=x^{3}-7x-1$$

#### Question. The diagram shows part of the curve $$y =\frac{1}{16}(3x – 1)^2$$ which touches the x-axis at the point P. The point Q (3, 4) lies on the curve and the tangent to the curve at Q crosses the x-axis at R.
Showing all necessary working, find by calculation
(i) State the x-coordinate of P.
(ii) the x-coordinate of R.
(iii) the area of the shaded region PQR.

Ans:(i) $$x=\frac{1}{3}$$

(ii) $$\frac{\partial y}{\partial x}=\left [ \frac{2}{16} (3x-1)\right ]\left [ 3\right ]$$
$$when x=3 n \to \frac{\partial y}{\partial x}=3$$
$$Equation or QR is y-4=3(x-3)$$
$$when y=0 \rightarrow x=\frac{5}{3}$$

(iii)Area under curve $$=\left [ \frac{1}{16\times 3}(3x-1)^3 \right ]\times \left [ \frac{1}{3} \right ]$$
$$\left [ \frac{1}{16\times 9}(8^3 -0)\right ]=\frac{32}{9}$$
Area of $$\Delta =\frac{8}{3}$$
Shaded area$$= \frac{32}{9}-\frac{8}{3}=\frac{8}{9}$$

### Question.

(a) Find  $$\int_{\infty }^{1} \frac{1}{\left ( 3x – 2 \right )^{\frac{3}{2}}}dx$$ The diagram shows the curve with equation y =$$\frac{1}{\left ( 3x-2 \right )^{\frac{3}{2}}}$$ . The shaded region is bounded by the curve, the x-axis and the lines x = 1 and x = 2. The shaded region is rotated through 3600 about the x-axis.

(b) Find the volume of revolution.

The normal to the curve at the point (1, 1) crosses the y-axis at the point A.

(c) Find the y-coordinate of A.

(a)  $$\left \{ \frac{\left ( 3x-2 \right )^{-\frac{1}{2}}}{-\frac{1}{2}} \right \}$$ ÷{3}

$$-\frac{2}{3}[0-1]$$

(b) $$\left [ \pi \right ]\int y^{2}dx = \left [ \pi \right ]\int \left ( 3x-2 \right )^{-3} dx = \left [ \pi \right ]\frac{\left ( 3x-2 \right )^{-2}}{-2\times 3}$$

$$\left [ \pi \right ]\left [ -\frac{1}{6} \right ]\left [ \frac{1}{16} -1\right ]$$

$$\frac{5\pi }{32}$$

(c) $$\frac{dy}{dx} = – \frac{3}{2}\times 3\left ( 3x-2 \right )^{\frac{5}{2}}$$

$$At x = 1, \frac{dy}{dx} = – \frac{9}{2}$$

[Equation of normal is] y – 1 = $$\frac{2}{9}$$ (x-1) OR evaluates c

At A, y = $$\frac{7}{9}$$