7 At the point \(( x , y )\), where \(x > 0\), the gradient of a curve is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { \frac { 1 } { 2 } } + \frac { 16 } { x ^ { 2 } } - 7$$
- Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 4\).
(1 mark) - Write \(\frac { 16 } { x ^ { 2 } }\) in the form \(16 x ^ { k }\), where \(k\) is an integer.
- Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
- Hence determine whether the point where \(x = 4\) is a maximum or a minimum, giving a reason for your answer.
- The point \(P ( 1,8 )\) lies on the curve.
- Show that the gradient of the curve at the point \(P\) is 12 .
- Find an equation of the normal to the curve at \(P\).
- Find \(\int \left( 3 x ^ { \frac { 1 } { 2 } } + \frac { 16 } { x ^ { 2 } } - 7 \right) \mathrm { d } x\).
- Hence find the equation of the curve which passes through the point \(P ( 1,8 )\).