(4 marks) d) Classify the extreme points of y(x)=2x^3-9x^2+12x e) Find (dy)/(dx) given (i) x^3-3x^2y^3=20(x+y^2) at x=-2,y=1 (4 marks) (ii) y=x^3cos^-1(3x) (3 marks) (iii) y=2^x^(2)ln(4x-14) (4 marks) (3 marks)

d) To classify the extreme points of the function $y(x)=2x^{3}-9x^{2}+12x$, we need to find the critical points by taking the derivative and setting it equal to zero.<br /><br />$y'(x) = 6x^2 - 18x + 12$<br /><br />Setting $y'(x) = 0$, we get:<br /><br />$6x^2 - 18x + 12 = 0$<br /><br />Simplifying, we have:<br /><br />$x^2 - 3x + 2 = 0$<br /><br />Factoring, we get:<br /><br />$(x-1)(x-2) = 0$<br /><br />So, the critical points are $x=1$ and $x=2$.<br /><br />To classify these critical points, we can use the second derivative test. Taking the second derivative of $y(x)$, we get:<br /><br />$y''(x) = 12x - 18$<br /><br />Evaluating $y''(x)$ at $x=1$, we get:<br /><br />$y''(1) = 12(1) - 18 = -6$<br /><br />Since $y''(1) < 0$, the critical point $x=1$ is a local maximum.<br /><br />Evaluating $y''(x)$ at $x=2$, we get:<br /><br />$y''(2) = 12(2) - 18 = 6$<br /><br />Since $y''(2) > 0$, the critical point $x=2$ is a local minimum.<br /><br />Therefore, the extreme points of the function $y(x)=2x^{3}-9x^{2}+12x$ are a local maximum at $x=1$ and a local minimum at $x=2$.<br /><br />e) (i) To find $\frac{dy}{dx}$ for the given equation $x^{3}-3x^{2}y^{3}=20(x+y^{2})$ at $x=-2,y=1$, we need to differentiate both sides of the equation with respect to $x$.<br /><br />Differentiating the left side, we get:<br /><br />$\frac{d}{dx}(x^3 - 3x^2y^3) = \frac{d}{dx}(20(x+y^2))$<br /><br />Simplifying, we have:<br /><br />$3x^2 - 3x^2y^3 \frac{dy}{dx} = 20 + 20y \frac{dy}{dx}$<br /><br />Rearranging the terms, we get:<br /><br />$3x^2 - 20 = 3x^2y^3 \frac{dy}{dx} - 20y \frac{dy}{dx}$<br /><br />Factoring out $\frac{dy}{dx}$, we have:<br /><br />$\frac{dy}{dx}(3x^2y^3 - 20y) = 3x^2 - 20$<br /><br />Solving for $\frac{dy}{dx}$, we get:<br /><br />$\frac{dy}{dx} = \frac{3x^2 - 20}{3x^2y^3 - 20y}$<br /><br />Substituting $x=-2$ and $y=1$ into the expression, we get:<br /><br />$\frac{dy}{dx} = \frac{3(-2)^2 - 20}{3(-2)^2(1)^3 - 20(1)} = \frac{12 - 20}{12 - 20} = \frac{-8}{-8} = 1$<br /><br />Therefore, $\frac{dy}{dx} = 1$ at $x=-2, y=1$.<br /><br />(ii) To find $\frac{dy}{dx}$ for the given function $y=x^{3}\cos^{-1}(3x)$, we need to use the chain rule and the derivative of the inverse trigonometric function.<br /><br />Differentiating $y$ with respect to $x$, we get:<br /><br />$\frac{dy}{dx} = \frac{d}{dx}(x^3) \cdot \cos^{-1}(3x) + x^3 \cdot \frac{d}{dx}(\cos^{-1}(3x))$<br /><br />Simplifying, we have:<br /><br />$\frac{dy}{dx} = 3x^2 \cdot \cos^{-1}(3x) + x^3 \cdot \frac{-1}{\sqrt{1-(3x)^2}} \cdot 3$<br /><br />Therefore, $\frac{dy}{dx} = 3x^2 \cos^{-1}(3x) - \frac{3x^3}{\sqrt{1-9x^2}}$.<br /><br />(iii) To find $\frac{dy}{dx}$ for the given function $y=2^{x^2}\ln(4x-14)$, we need to use the chain rule and the derivative of the exponential and logarithmic functions.<br /><br />Differentiating $y$ with