Find stationary points coordinates

8 A curve has equation \(y = f ( x )\) where \(f ( x ) = \frac { 4 x ^ { 3 } + 8 x - 4 } { 2 x - 1 }\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the coordinates of each of the stationary points of the curve \(y = \mathrm { f } ( x )\).
2. Divide \(4 x ^ { 3 } + 8 x - 4\) by ( \(2 x - 1\) ), and hence find \(\int \mathrm { f } ( x ) \mathrm { d } x\).
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10 \includegraphics[max width=\textwidth, alt={}, center]{b6bede75-3da4-4dda-9303-a5a692fc2572-3_556_1093_1596_523} The diagram shows the curve \(y = 10 e ^ { - \frac { 1 } { 2 } x } \sin 4 x\) for \(x \geqslant 0\). The stationary points are labelled \(T _ { 1 } , T _ { 2 }\), \(T _ { 3 } , \ldots\) as shown.

1. Find the \(x\)-coordinates of \(T _ { 1 }\) and \(T _ { 2 }\), giving each \(x\)-coordinate correct to 3 decimal places.
2. It is given that the \(x\)-coordinate of \(T _ { n }\) is greater than 25 . Find the least possible value of \(n\).

4 The curve with equation \(y = \frac { \mathrm { e } ^ { 2 x } } { 4 + \mathrm { e } ^ { 3 x } }\) has one stationary point. Find the exact values of the coordinates of this point.

3 The curve \(y = \frac { \mathrm { e } ^ { x } } { \cos x }\), for \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\), has one stationary point. Find the \(x\)-coordinate of this point.

2 The curve with equation \(y = \frac { \mathrm { e } ^ { - 2 x } } { 1 - x ^ { 2 } }\) has a stationary point in the interval \(- 1 < x < 1\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of this stationary point, giving the answer correct to 3 decimal places.

3 Find the exact coordinates of the stationary point of the curve with equation \(y = \frac { 3 x } { \ln x }\).

2 The curve with equation \(y = \frac { \sin 2 x } { \mathrm { e } ^ { 2 x } }\) has one stationary point in the interval \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Find the exact \(x\)-coordinate of this point.

2 The curve \(y = \frac { \mathrm { e } ^ { 3 x - 1 } } { 2 x }\) has one stationary point. Find the coordinates of this stationary point.

5 Find the \(x\)-coordinates of the stationary points of the following curves:

1. \(y = 4 x \mathrm { e } ^ { - 3 x }\);
2. \(y = \frac { 4 x ^ { 2 } } { x + 1 }\).

5 Find the exact coordinates of the stationary point of the curve with equation \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } ( 2 x + 5 )\).

8 The equation of a curve is \(y = e ^ { - 5 x } \tan ^ { 2 } x\) for \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\).
Find the \(x\)-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate.

4 The equation of a curve is \(y = \cos ^ { 3 } x \sqrt { \sin x }\). It is given that the curve has one stationary point in the interval \(0 < x < \frac { 1 } { 2 } \pi\). Find the \(x\)-coordinate of this stationary point, giving your answer correct to 3 significant figures.

5 The equation of a curve is \(y = \frac { e ^ { \sin x } } { \cos ^ { 2 } x }\) for \(0 \leqslant x \leqslant 2 \pi\).
Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) and hence find the \(x\)-coordinates of the stationary points of the curve.

3 The equation of a curve is \(y = \sin x \sin 2 x\). The curve has a stationary point in the interval \(0 < x < \frac { 1 } { 2 } \pi\). Find the \(x\)-coordinate of this point, giving your answer correct to 3 significant figures.

5 Find the exact coordinates of the stationary points of the curve \(y = \frac { \mathrm { e } ^ { 3 x ^ { 2 } - 1 } } { 1 - x ^ { 2 } }\).

9. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. A curve has equation $$y = \frac { 4 x ^ { 2 } + 9 } { 2 \sqrt { x } } \quad x > 0$$ Find the \(x\) coordinate of the point on the curve at which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\)

4. (i) The curve \(C\) has equation $$y = \frac { x } { 9 + x ^ { 2 } }$$ Use calculus to find the coordinates of the turning points of \(C\).
(ii) Given that $$y = \left( 1 + \mathrm { e } ^ { 2 x } \right) ^ { \frac { 3 } { 2 } }$$ find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = \frac { 1 } { 2 } \ln 3\).

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with the equation \(y = \left( 2 x ^ { 2 } - 5 x + 2 \right) \mathrm { e } ^ { - x }\).

1. Find the coordinates of the point where \(C\) crosses the \(y\)-axis.
2. Show that \(C\) crosses the \(x\)-axis at \(x = 2\) and find the \(x\)-coordinate of the other point where \(C\) crosses the \(x\)-axis.
3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
4. Hence find the exact coordinates of the turning points of \(C\).

1. (i) (a) Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x ^ { \frac { 1 } { 2 } } \ln x \right) = \frac { \ln x } { 2 \sqrt { } x } + \frac { 1 } { \sqrt { } x }\)

The curve with equation \(y = x ^ { \frac { 1 } { 2 } } \ln x , x > 0\) has one turning point at the point \(P\).
(b) Find the exact coordinates of \(P\). Give your answer in its simplest form.
(ii) A curve \(C\) has equation \(y = \frac { x - k } { x + k }\), where \(k\) is a positive constant. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), and show that \(C\) has no turning points.

6 A curve has equation \(y = \frac { x } { 2 + 3 \ln x }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence find the exact coordinates of the stationary point of the curve.

3. Find the coordinates of the stationary points of the curve with equation $$y = \frac { x - 1 } { x ^ { 2 } - 2 x + 5 }$$

5 The equation of a curve is \(y = \frac { x ^ { 2 } } { 2 x + 1 }\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ( x + 1 ) } { ( 2 x + 1 ) ^ { 2 } }\).
2. Find the coordinates of the stationary points of the curve. You need not determine their nature.

7 Fig. 7 shows the graphs of the curves \(y = \mathrm { e } ^ { - x }\) and \(y = \mathrm { e } ^ { - x } \sin x\) for \(0 \leq x \leq \pi\). \begin{figure}[h] \end{figure} The maximum point on \(y = \mathrm { e } ^ { - x } \sin x\) is at A , and the curves touch at B . \(\mathrm { A } ^ { \prime }\) and \(\mathrm { B } ^ { \prime }\) are the points on the \(x\)-axis such that \(\mathrm { A } ^ { \prime } \mathrm { A }\) and \(\mathrm { B } ^ { \prime } \mathrm { B }\) are parallel to the \(y\)-axis.
Show that \(\mathrm { OA } ^ { \prime } = \mathrm { A } ^ { \prime } \mathrm { B } ^ { \prime }\).

1 Fig. 1 shows part of the curve \(y = \mathrm { e } ^ { 2 x } \cos x\). \begin{figure}[h] \end{figure} Find the coordinates of the turning point P .