Derive stationary point equation

A question is this type if and only if it asks to show using calculus that the x-coordinate of a stationary/turning point satisfies a particular equation.

32 questions · Standard +0.4

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Edexcel C34 2014 January Q7
11 marks Standard +0.3
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5b698944-41ac-4072-b5e1-c580b7752c39-20_689_712_248_680} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = 2 x ( 1 + x ) \ln x , \quad x > 0$$ The curve has a minimum turning point at \(A\).
  1. Find f'(x)
  2. Hence show that the \(x\) coordinate of \(A\) is the solution of the equation $$x = \mathrm { e } ^ { - \frac { 1 + x } { 1 + 2 x } }$$
  3. Use the iteration formula $$x _ { n + 1 } = \mathrm { e } ^ { - \frac { 1 + x _ { n } } { 1 + 2 x _ { n } } } , \quad x _ { 0 } = 0.46$$ to find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 4 decimal places.
  4. Use your answer to part (c) to estimate the coordinates of \(A\) to 2 decimal places.
Edexcel C34 2015 January Q10
10 marks Standard +0.3
10. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-17_598_736_223_603} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 4 , \quad x > 0$$ Point \(A\) is the minimum turning point on the curve.
  1. Show, by using calculus, that the \(x\) coordinate of point \(A\) is a solution of $$x = \frac { 6 } { 1 + \ln \left( x ^ { 2 } \right) }$$
  2. Starting with \(x _ { 0 } = 2.27\), use the iteration $$x _ { n + 1 } = \frac { 6 } { 1 + \ln \left( x _ { n } ^ { 2 } \right) }$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
  3. Use your answer to part (b) to deduce the coordinates of point \(A\) to one decimal place.
Edexcel C3 2012 January Q6
12 marks Standard +0.3
6. $$f ( x ) = x ^ { 2 } - 3 x + 2 \cos \left( \frac { 1 } { 2 } x \right) , \quad 0 \leqslant x \leqslant \pi$$
  1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a solution in the interval \(0.8 < x < 0.9\) The curve with equation \(y = \mathrm { f } ( x )\) has a minimum point \(P\).
  2. Show that the \(x\)-coordinate of \(P\) is the solution of the equation $$x = \frac { 3 + \sin \left( \frac { 1 } { 2 } x \right) } { 2 }$$
  3. Using the iteration formula $$x _ { n + 1 } = \frac { 3 + \sin \left( \frac { 1 } { 2 } x _ { n } \right) } { 2 } , \quad x _ { 0 } = 2$$ find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
  4. By choosing a suitable interval, show that the \(x\)-coordinate of \(P\) is 1.9078 correct to 4 decimal places.
Edexcel C3 2005 June Q4
9 marks Moderate -0.3
4. $$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$
  1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
  2. Show that \(\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }\). The iterative formula $$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , x _ { 0 } = 1$$ is used to find an approximate value for \(\alpha\).
  3. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
  4. By considering the change of sign of \(\mathrm { f } ^ { \prime } ( x )\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places.
Edexcel C3 2006 June Q5
11 marks Standard +0.3
5. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{f0f328ed-3550-4b8d-8b80-016df8773b21-07_465_565_296_701}
\end{figure} Figure 2 shows part of the curve with equation $$y = ( 2 x - 1 ) \tan 2 x , \quad 0 \leqslant x < \frac { \pi } { 4 }$$ The curve has a minimum at the point \(P\). The \(x\)-coordinate of \(P\) is \(k\).
  1. Show that \(k\) satisfies the equation $$4 k + \sin 4 k - 2 = 0$$ The iterative formula $$x _ { n + 1 } = \frac { 1 } { 4 } \left( 2 - \sin 4 x _ { n } \right) , x _ { 0 } = 0.3$$ is used to find an approximate value for \(k\).
  2. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
  3. Show that \(k = 0.277\), correct to 3 significant figures.
OCR C3 Q8
13 marks Standard +0.3
  1. A curve has the equation \(y = x ^ { 2 } - \sqrt { 4 + \ln x }\).
    1. Show that the tangent to the curve at the point where \(x = 1\) has the equation
    $$7 x - 4 y = 11$$ The curve has a stationary point with \(x\)-coordinate \(\alpha\).
  2. Show that \(0.3 < \alpha < 0.4\)
  3. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
  4. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$ with \(x _ { 0 } = 0.35\), to find \(\alpha\) correct to 5 decimal places.
    You should show the result of each iteration.
OCR C3 2011 January Q6
9 marks Standard +0.8
6 The curve with equation \(y = \frac { 3 x + 4 } { x ^ { 3 } - 4 x ^ { 2 } + 2 }\) has a stationary point at \(P\). It is given that \(P\) is close to the point with coordinates \(( 2.4 , - 1.6 )\).
  1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \sqrt [ 3 ] { \frac { 16 } { 3 } x + 1 }$$
  2. By first using an iterative process based on the equation in part (i), find the coordinates of \(P\), giving each coordinate correct to 3 decimal places.