Prove curve has no turning points

A question is this type if and only if it asks to prove or show that a given curve has no stationary/turning points, typically by showing the discriminant of the derivative is negative.

6 questions · Moderate -0.1

1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives
Sort by: Default | Easiest first | Hardest first
Edexcel P2 2019 October Q10
7 marks Moderate -0.3
10. The curve \(C\) has equation $$y = a x ^ { 3 } - 3 x ^ { 2 } + 3 x + b$$ where \(a\) and \(b\) are constants. Given that
  • the point \(( 2,5 )\) lies on \(C\)
  • the gradient of the curve at \(( 2,5 )\) is 7
    1. find the value of \(a\) and the value of \(b\).
    2. Prove that \(C\) has no turning points.
OCR MEI AS Paper 2 2023 June Q11
5 marks Moderate -0.8
11 In this question you must show detailed reasoning.
The equation of a curve is \(y = 2 x ^ { 3 } + 9 x ^ { 2 } + 24 x - 8\).
Show that there are no stationary points on this curve.
OCR MEI Paper 2 2024 June Q8
6 marks Standard +0.8
8 The equation of a curve is \(y = 2 x ^ { 3 } + 3 m x ^ { 2 } - 9 m x + 4\). Determine the range of values of \(m\) for which the curve has no stationary values.
OCR C1 2006 June Q6
8 marks Moderate -0.3
  1. Solve the equation \(x^4 - 10x^2 + 25 = 0\). [4]
  2. Given that \(y = \frac{2}{5}x^5 - \frac{20}{3}x^3 + 50x + 3\), find \(\frac{dy}{dx}\). [2]
  3. Hence find the number of stationary points on the curve \(y = \frac{2}{5}x^5 - \frac{20}{3}x^3 + 50x + 3\). [2]
AQA AS Paper 1 2019 June Q8
6 marks Standard +0.3
Prove that the curve with equation $$y = 2x^5 + 5x^4 + 10x^3 - 8$$ has only one stationary point, stating its coordinates. [6 marks]
SPS SPS SM 2022 February Q2
8 marks Moderate -0.3
  1. Solve the equation \(x^4 - 10x^2 + 25 = 0\). [4]
  2. Given that \(y = \frac{2}{5}x^5 - \frac{20}{3}x^3 + 50x + 3\), find \(\frac{dy}{dx}\). [2]
  3. Hence find the number of stationary points on the curve \(y = \frac{2}{5}x^5 - \frac{20}{3}x^3 + 50x + 3\). [2]