The turning points of the curve are at the points where the gradient is zero.

Find the gradient of the curve in terms of $x$ by differentiating.

Formula for differentiation: If $\begin{array}{rcl} y & = & a x^{b} \end{array}$ , $\begin{array}{rcl} \frac{d y}{d x} & = & a b x^{b - 1} \end{array}$ .

To differentiate the first term, $x^{3}$ .

$\begin{array}{rcl} y & = & x^{3} \\ \frac{d y}{d x} & = & 3 x^{3 - 1} = 3 x^{2} \end{array}$

[1]

To differentiate the second term, $- 6 x^{2}$ .

$\begin{array}{rcl} y & = & - 6 x^{2} \\ \frac{d y}{d x} & = & (- 6 \times 2) x^{2 - 1} = - 12 x \end{array}$

[1]

To differentiate the third term, $16$ .

$\begin{array}{rcl} y & = & 16 x^{0} \\ \frac{d y}{d x} & = & (16 \times 0) x^{0 - 1} = 0 \end{array}$

Remember, the derivative of a constant term will always by zero.
To find the turning points, set $\frac{d y}{d x}$ equal to zero.

$\begin{array}{rcl} \frac{d y}{d x} & = & 0 \\ 3 x^{2} - 12 x & = & 0 \end{array}$

[1]

Use a method to solve the quadratic, factorising is easiest here.

$\begin{array}{rcl} 3 x (x - 4^{}) & = & 0 \end{array}$

[1]

Solve by setting each part to zero.

$\begin{array}{rcl} 3 x & = & 0 \\ x_{1} & = & 0 \end{array}$

and,

$\begin{array}{rcl} (x - 4) & = & 0 \\ x_{2} & = & 4 \end{array}$

Find the $y$ coordinate by substituting the values for $x$ into the equation for the curve.

$\begin{array}{rcl} {y_{1}}^{} & = & {x_{1}}^{3} - 6 {x_{1}}^{2} + 16 \\ = & {(0)}^{3} - 6 {(0)}^{2} + 16 \\ = & 16 \end{array}$

and,

$\begin{array}{rcl} {y_{2}}^{} & = & {x_{2}}^{3} - 6 {x_{2}}^{2} + 16 \\ = & {(4)}^{3} - 6 {(4)}^{2} + 16 \\ = & - 16 \end{array}$

$\begin{array}{rcl} (0, 16) and (4, - 16) \end{array}$ [1]

Differentiation (CIE IGCSE Maths: Extended)

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