Stationary points and nature classification

A question is this type if and only if it requires finding stationary points of a given function and determining whether each is a maximum or minimum using the second derivative or sign change, without an optimisation constraint.

2 questions · Moderate -0.8

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OCR H240/03 2018 September Q1
5 marks Easy -1.8
1
  1. Show that \(4 x ^ { 2 } - 12 x + 3 = 4 \left( x - \frac { 3 } { 2 } \right) ^ { 2 } - 6\).
  2. State the coordinates of the minimum point of the curve \(y = 4 x ^ { 2 } - 12 x + 3\).
AQA C3 2015 June Q3
14 marks Standard +0.3
3
  1. It is given that the curves with equations \(y = 6 \ln x\) and \(y = 8 x - x ^ { 2 } - 3\) intersect at a single point where \(x = \alpha\).
    1. Show that \(\alpha\) lies between 5 and 6 .
    2. Show that the equation \(x = 4 + \sqrt { 13 - 6 \ln x }\) can be rearranged into the form $$6 \ln x + x ^ { 2 } - 8 x + 3 = 0$$
    3. Use the iterative formula $$x _ { n + 1 } = 4 + \sqrt { 13 - 6 \ln x _ { n } }$$ with \(x _ { 1 } = 5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
  2. A curve has equation \(y = \mathrm { f } ( x )\) where \(\mathrm { f } ( x ) = 6 \ln x + x ^ { 2 } - 8 x + 3\).
    1. Find the exact values of the coordinates of the stationary points of the curve.
    2. Hence, or otherwise, find the exact values of the coordinates of the stationary points of the curve with equation $$y = 2 \mathrm { f } ( x - 4 )$$