The curve C has equation $y = 12ig(\sqrt{x}\big) - x^{\frac{3}{2}} - 10,$ $x > 0$ (a) Use calculus to find the coordinates of the turning point on C - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 4

To find the second derivative, first differentiate the first derivative:

1. The first derivative is:
   [ \frac{dy}{dx} = \frac{6}{\sqrt{x}} - \frac{3}{2}x^{\frac{1}{2}} ]

2. Differentiate again:
   [ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{6}{\sqrt{x}}) - \frac{d}{dx}(\frac{3}{2}x^{\frac{1}{2}}) ] The derivative of the first term is: [ -3x^{-\frac{3}{2}} ]
   And the derivative of the second term is: [ \frac{3}{4}x^{-\frac{1}{2}} ]

Combining these results: [ \frac{d^{2}y}{dx^{2}} = -3x^{-\frac{3}{2}} - \frac{3}{4}x^{-\frac{1}{2}} ]

At the turning point, we found that $x = 4$. To determine the nature of the turning point, we will evaluate the second derivative:

1. Substitute $x = 4$ into $\frac{d^{2}y}{dx^{2}}$: [ \frac{d^{2}y}{dx^{2}} = -3(4^{-\frac{3}{2}}) - \frac{3}{4}(4^{-\frac{1}{2}}) ] Calculating these values gives: [ \frac{d^{2}y}{dx^{2}} < 0 ]

2. Since the second derivative is negative, this indicates that the turning point is a maximum.

Thus, the nature of the turning point is:
[\text{It is a maximum.}]