e.g. Find the equation of the tangent to the curve $y = x^3 + 6x - 4$ when $x = 1$.

1. Find the gradient function, $\frac{dy}{dx}$ , of the curve.

$\frac{dy}{dx} = 3x^2 + 6 \Rightarrow \text{When } x = 1, \, \frac{dy}{dx} = 3(1)^2 + 6 = 9$

2. Find both coordinates of the point in question.

$x = 1 \Rightarrow y = (1)^3 + 6(1) - 4 = 3$

3. Use this point and gradient in the general straight line equation.

$m = 9 \quad \text{Point } (1,3)$

$\therefore y - 3 = 9(x - 1) \Rightarrow y - 3 = 9x - 9 \Rightarrow \boxed{y = 9x - 6}$

e.g. Find the equation of the normal to $y = 4x^2 + x - 3$ when $x = 2$

$\frac{dy}{dx} = 8x + 1 = 8(2) + 1 = 17 \quad (\text{when } x = 2)$

$\Rightarrow \text{Gradient of normal} = -\frac{1}{17}$

$y = 4(2)^2 + 2 - 3 = 15$

$\therefore \, m = -\frac{1}{17} \quad \text{Point } (2, 15)$

$\Rightarrow y - 15 = -\frac{1}{17}(x - 2)$

$\Rightarrow 17y - 255 = -1(x - 2) \quad \left\{ \times 17 \right\}$

$\Rightarrow 17y - 255 = -x + 2$

$\Rightarrow x + 17y - 257 = 0$

Solution:

(i) $\frac{dy}{dx} = 3x^2 - 6 \checkmark$

(ii)$3x^2 - 6 < 0$

$\Rightarrow x^2 - 2 < 0$

$\Rightarrow (x + \sqrt{2})(x - \sqrt{2}) < 0$

$\Rightarrow -\sqrt{2} < x < \sqrt{2} \checkmark$

(Diagram illustrating the interval between $-\sqrt{2}$ and $\sqrt{2}$ on the x-axis

(iii) Let $x = -1$:

$\Rightarrow \frac{dy}{dx} = 3(-1)^2 - 6 = -3$

$\text{Point } (-1, 7) \quad m = -3$

$\Rightarrow y - 7 = -3(x - 1)$

$\Rightarrow y - 7 = -3x - 3$

$\Rightarrow y = -3x + 4$

$-3x + 4 = x^3 - 6x + 2 \Rightarrow x^3 - 3x - 2 = 0$

$\Rightarrow x = 2 \, (\text{valid root})$

$\Rightarrow y = (2)^3 - 6(2) + 2 = -2 \quad \Rightarrow (2, -2) \checkmark$

$f(x) = 6x^{\frac{5}{2}} + 4 \Rightarrow f'(x) = \frac{5}{2} \times 6x^{\frac{3}{2}}$

$= 15x^{\frac{3}{2}}$