1 Find the range of values of $$ x $$ for which (a) $$ y=x^{2}+2 x-4 $$ is an increasing function, (b) $$ y=3 x^{2}+4 x-3 $$ is a decreasing function. The equation of a curve is $$ y=\frac{1}{9}(3 x-1)^{3}-3 x $$. (i) Find the gradient of the normal at the point $$ P $$ where the curve intersects the $$ y $$-axis. (ii) Find the equation of the normal at $$ P $$ in the form $$ p x+q y+r=0 $$, where $$ p, q $$ and $$ r $$ are integers. (iii) The normal at $$ P $$ meets the $$ x $$-axis at the point $$ Q $$. Find the area of triangle $$ O P Q $$. (iv) Find the range of values of $$ x $$ for which $$ y=\frac{1}{9}(3 x-1)^{3}-3 x $$ is an increasing function of $$ x $$.

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**(a)** For the function $$ y=x^2+2x-4, $$ its derivative is $$ \frac{dy}{dx}=2x+2. $$ The function is increasing when $$ 2x+2\ge0. $$ Dividing by 2, $$ x+1\ge0\quad\Longrightarrow\quad x\ge-1. $$ --- **(b)** For the function $$ y=3x^2+4x-3, $$ its derivative is $$ \frac{dy}{dx}=6x+4. $$ The function is decreasing when $$ 6x+4<0. $$ Solving, $$ 6x<-4\quad\Longrightarrow\quad x<-\frac{2}{3}. $$ --- The curve is given by $$ y=\frac{1}{9}(3x-1)^3-3x. $$ **(i)** The curve meets the $$y$$-axis when $$x=0$$. Substituting $$x=0$$ into the equation gives $$ y=\frac{1}{9}(3\cdot0-1)^3-3\cdot0=\frac{1}{9}(-1)^3=-\frac{1}{9}. $$ So the point $$P$$ is $$ P=\left(0,-\frac{1}{9}\right). $$ Differentiate $$y$$ with respect to $$x$$. Writing $$ y=\frac{1}{9}(3x-1)^3-3x, $$ let $$u=3x-1$$. Then $$ \frac{dy}{dx}=\frac{1}{9}\cdot 3u^2\cdot\frac{du}{dx}-3, $$ and since $$\frac{du}{dx}=3$$, we have $$ \frac{dy}{dx}=\frac{1}{9}\cdot3(3x-1)^2\cdot3-3=(3x-1)^2-3. $$ At $$x=0$$: $$ \frac{dy}{dx}=(3\cdot0-1)^2-3=( -1)^2-3=1-3=-2. $$ This is the tangent gradient. The gradient of the normal is the negative reciprocal of the tangent gradient: $$ m_{\text{normal}}=-\frac{1}{-2}=\frac{1}{2}. $$ **(ii)** The normal at $$P=(0,-\frac{1}{9})$$ with gradient $$\frac{1}{2}$$ has the point-slope form $$ y-\left(-\frac{1}{9}\right)=\frac{1}{2}(x-0), $$ which simplifies to $$ y+\frac{1}{9}=\frac{1}{2}x. $$ Multiplying through by 18 to clear fractions gives: $$ 18y+2=9x. $$ Rearranging into the form $$px+qy+r=0$$: $$ 9x-18y-2=0. $$ **(iii)** The normal meets the $$x$$-axis at point $$Q$$ (where $$y=0$$). Substitute $$y=0$$ into the equation of the normal: $$ 0+\frac{1}{9}=\frac{1}{2}x \quad\Longrightarrow\quad \frac{1}{9}=\frac{x}{2}. $$ Thus, $$ x=\frac{2}{9}. $$ So, $$Q=\left(\frac{2}{9},0\right)$$. We wish to find the area of triangle $$OPQ$$ where $$O=(0,0)$$, $$P=\left(0,-\frac{1}{9}\right)$$, and $$Q=\left(\frac{2}{9},0\right)$$. The triangle is a right triangle with base $$ \text{base}=\frac{2}{9}\quad \text{(along the $$x$$-axis)} $$ and height $$ \text{height}=\frac{1}{9}\quad \text{(vertical distance)}. $$ Therefore, its area is given by $$ \text{Area}=\frac{1}{2}\times\frac{2}{9}\times\frac{1}{9}=\frac{1}{81}. $$ **(iv)** For the function $$ y=\frac{1}{9}(3x-1)^3-3x, $$ we already found that $$ \frac{dy}{dx}=(3x-1)^2-3. $$ The function is increasing when $$ (3x-1)^2-3>0. $$ That is, $$ (3x-1)^2>3. $$ Taking square roots (and remembering the absolute value), we have $$ \lvert3x-1\rvert>\sqrt{3}. $$ This gives the two conditions: - $$3x-1>\sqrt{3}$$ which implies $$ 3x>1+\sqrt{3}\quad\Longrightarrow\quad x>\frac{1+\sqrt{3}}{3}, $$ - $$3x-1<-\sqrt{3}$$ which implies $$ 3x<1-\sqrt{3}\quad\Longrightarrow\quad x<\frac{1-\sqrt{3}}{3}. $$ Therefore, the function is increasing when $$ x<\frac{1-\sqrt{3}}{3}\quad \text{or}\quad x>\frac{1+\sqrt{3}}{3}. $$ **(a)** The function $$ y = x^2 + 2x - 4 $$ is increasing when $$ x \geq -1 $$. **(b)** The function $$ y = 3x^2 + 4x - 3 $$ is decreasing when $$ x < -\frac{2}{3} $$. **(i)** The gradient of the normal at point $$ P $$ is $$ \frac{1}{2} $$. **(ii)** The equation of the normal at $$ P $$ is $$ 9x - 18y - 2 = 0 $$. **(iii)** The area of triangle $$ OPQ $$ is $$ \frac{1}{81} $$. **(iv)** The function $$ y = \frac{1}{9}(3x-1)^3 - 3x $$ is increasing when $$ x < \frac{1 - \sqrt{3}}{3} $$ or $$ x > \frac{1 + \sqrt{3}}{3} $$.

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