1.07n Stationary points: find maxima, minima using derivatives

8

1. Find the coordinates of the stationary point on the curve \(y = 3 x ^ { 2 } - \frac { 6 } { x } - 2\).
2. Determine whether the stationary point is a maximum point or a minimum point.

10 A curve has equation \(y = ( 2 x - 1 ) ( x + 3 ) ( x - 1 )\).

1. Sketch the curve, indicating the coordinates of all points of intersection with the axes.
2. Show that the gradient of the curve at the point \(P ( 1,0 )\) is 4 .
3. The line \(l\) is parallel to the tangent to the curve at the point \(P\). The curve meets \(l\) at the point where \(x = - 2\). Find the equation of \(l\), giving your answer in the form \(y = m x + c\).
4. Determine whether \(l\) is a tangent to the curve at the point where \(x = - 2\).

8

1. Find the coordinates of the stationary point on the curve \(y = x ^ { 4 } + 32 x\).
2. Determine whether this stationary point is a maximum or a minimum.
3. For what values of \(x\) does \(x ^ { 4 } + 32 x\) increase as \(x\) increases?

9 The curve \(y = 2 x ^ { 3 } - a x ^ { 2 } + 8 x + 2\) passes through the point \(B\) where \(x = 4\).

1. Given that \(B\) is a stationary point of the curve, find the value of the constant \(a\).
2. Determine whether the stationary point \(B\) is a maximum point or a minimum point.
3. Find the \(x\)-coordinate of the other stationary point of the curve.

11 The curve \(y = 4 x ^ { 2 } + \frac { a } { x } + 5\) has a stationary point. Find the value of the positive constant \(a\) given that the \(y\)-coordinate of the stationary point is 32 .

12 The curve with equation \(y = \frac { 1 } { 5 } x ( 10 - x )\) is used to model the arch of a bridge over a road, where \(x\) and \(y\) are distances in metres, with the origin as shown in Fig. 12.1. The \(x\)-axis represents the road surface. \begin{figure}[h] \end{figure}

1. State the value of \(x\) at A , where the arch meets the road.
2. Using symmetry, or otherwise, state the value of \(x\) at the maximum point B of the graph. Hence find the height of the arch.
3. Fig. 12.2 shows a lorry which is 4 m high and 3 m wide, with its cross-section modelled as a rectangle. Find the value of \(d\) when the lorry is in the centre of the road. Hence show that the lorry can pass through this arch. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ede57eaa-2645-49df-aa09-68b6d5f35a9a-4_529_871_1489_678} \captionsetup{labelformat=empty} \caption{Fig. 12.2}
   \end{figure}
4. Another lorry, also modelled as having a rectangular cross-section, has height 4.5 m and just touches the arch when it is in the centre of the road. Find the width of this lorry, giving your answer in surd form.
   [0pt] [5]

7 Differentiate \(4 x ^ { 2 } + \frac { 1 } { x }\) and hence find the \(x\)-coordinate of the stationary point of the curve \(y = 4 x ^ { 2 } + \frac { 1 } { x }\).

12 The equation of a curve is \(y = 9 x ^ { 2 } - x ^ { 4 }\).

1. Show that the curve meets the \(x\)-axis at the origin and at \(x = \pm a\), stating the value of \(a\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\). Hence show that the origin is a minimum point on the curve. Find the \(x\)-coordinates of the maximum points.
3. Use calculus to find the area of the region bounded by the curve and the \(x\)-axis between \(x = 0\) and \(x = a\), using the value you found for \(a\) in part (i).

11

1. The standard formulae for the volume \(V\) and total surface area \(A\) of a solid cylinder of radius \(r\) and height \(h\) are $$V = \pi r ^ { 2 } h \quad \text { and } \quad A = 2 \pi r ^ { 2 } + 2 \pi r h .$$ Use these to show that, for a cylinder with \(A = 200\), $$V = 100 r - \pi r ^ { 3 }$$
2. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) and \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} r ^ { 2 } }\).
3. Use calculus to find the value of \(r\) that gives a maximum value for \(V\) and hence find this maximum value, giving your answers correct to 3 significant figures.

10

1. Use calculus to find, correct to 1 decimal place, the coordinates of the turning points of the curve \(y = x ^ { 3 } - 5 x\). [You need not determine the nature of the turning points.]
2. Find the coordinates of the points where the curve \(y = x ^ { 3 } - 5 x\) meets the axes and sketch the curve.
3. Find the equation of the tangent to the curve \(y = x ^ { 3 } - 5 x\) at the point \(( 1 , - 4 )\). Show that, where this tangent meets the curve again, the \(x\)-coordinate satisfies the equation $$x ^ { 3 } - 3 x + 2 = 0$$ Hence find the \(x\)-coordinate of the point where this tangent meets the curve again.

10 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x + 3\). The curve passes through the point ( 2,9 ).

1. Find the equation of the tangent to the curve at the point \(( 2,9 )\).
2. Find the equation of the curve and the coordinates of its points of intersection with the \(x\)-axis. Find also the coordinates of the minimum point of this curve.
3. Find the equation of the curve after it has been stretched parallel to the \(x\)-axis with scale factor \(\frac { 1 } { 2 }\). Write down the coordinates of the minimum point of the transformed curve.

4 For each of the following curves, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and determine the exact \(x\)-coordinate of the stationary point:

1. \(y = \left( 4 x ^ { 2 } + 1 \right) ^ { 5 }\),
2. \(y = \frac { x ^ { 2 } } { \ln x }\).

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.

9 \includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-4_661_915_269_557} The diagram shows the curve $$y = \mathrm { e } ^ { 2 x } - 18 x + 15 .$$ The curve crosses the \(y\)-axis at \(P\) and the minimum point is \(Q\). The shaded region is bounded by the curve and the line \(P Q\).

1. Show that the \(x\)-coordinate of \(Q\) is \(\ln 3\).
2. Find the exact area of the shaded region.

8 \includegraphics[max width=\textwidth, alt={}, center]{33a2b09d-0df9-48d6-9ee9-e0a1ec345f41-4_616_1024_296_516} The diagram shows the curve \(y = \frac { 2 x + 4 } { x ^ { 2 } + 5 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the coordinates of the two stationary points.
2. The function g is defined for all real values of \(x\) by $$\mathrm { g } ( x ) = \left| \frac { 2 x + 4 } { x ^ { 2 } + 5 } \right| .$$
   1. Sketch the curve \(y = \mathrm { g } ( x )\) and state the range of g .
   2. It is given that the equation \(\mathrm { g } ( x ) = k\), where \(k\) is a constant, has exactly two distinct real roots. Write down the set of possible values of \(k\).

5 \includegraphics[max width=\textwidth, alt={}, center]{00a4be37-c095-4d9c-a1cd-d03b8ab1d411-2_455_643_1327_694} The diagram shows the curve \(y = \mathrm { e } ^ { 3 x } - 6 \mathrm { e } ^ { 2 x } + 32\).

1. Find the exact \(x\)-coordinate of the minimum point and verify that the \(y\)-coordinate of the minimum point is 0 .
2. Find the exact area of the region (shaded in the diagram) enclosed by the curve and the axes.

8 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\). \begin{figure}[h] \end{figure}

1. Find the gradient of PR.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
3. Find the exact coordinates of the turning point Q .
4. Differentiate \(x \ln x - x\). Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 x - x ^ { 2 } } }\).
The curve has asymptotes \(x = 0\) and \(x = a\). \begin{figure}[h] \end{figure}

1. Find \(a\). Hence write down the domain of the function.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 1 } { \left( 2 x - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the coordinates of the turning point of the curve, and write down the range of the function. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
3. (A) Show algebraically that \(\mathrm { g } ( x )\) is an even function.
   (B) Show that \(\mathrm { g } ( x - 1 ) = \mathrm { f } ( x )\).
   (C) Hence prove that the curve \(y = \mathrm { f } ( x )\) is symmetrical, and state its line of symmetry.

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 1 } { x ^ { 2 } + 1 }\) for the domain \(0 \leqslant x \leqslant 2\). \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }\), and hence that \(\mathrm { f } ( x )\) is an increasing function for \(x > 0\).
2. Find the range of \(\mathrm { f } ( x )\).
3. Given that \(\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 6 - 18 x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 3 } }\), find the maximum value of \(\mathrm { f } ^ { \prime } ( x )\). The function \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
4. Write down the domain and range of \(\mathrm { g } ( x )\). Add a sketch of the curve \(y = \mathrm { g } ( x )\) to a copy of Fig. 9 .
5. Show that \(\mathrm { g } ( x ) = \sqrt { \frac { x + 1 } { 2 - x } }\).

9 Fig. 9 shows the curve \(y = \frac { x ^ { 2 } } { 3 x - 1 }\).
P is a turning point, and the curve has a vertical asymptote \(x = a\). \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ( 3 x - 2 ) } { ( 3 x - 1 ) ^ { 2 } }\).
3. Find the exact coordinates of the turning point P . Calculate the gradient of the curve when \(x = 0.6\) and \(x = 0.8\), and hence verify that P is a minimum point.
4. Using the substitution \(u = 3 x - 1\), show that \(\int \frac { x ^ { 2 } } { 3 x - 1 } \mathrm {~d} x = \frac { 1 } { 27 } \int \left( u + 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = \frac { 2 } { 3 }\) and \(x = 1\).

8 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = ( 1 - x ) \mathrm { e } ^ { 2 x }\), with its turning point P . \begin{figure}[h] \end{figure}

1. Write down the coordinates of the intercepts of \(y = \mathrm { f } ( x )\) with the \(x\) - and \(y\)-axes.
2. Find the exact coordinates of the turning point P .
3. Show that the exact area of the region enclosed by the curve and the \(x\) - and \(y\)-axes is \(\frac { 1 } { 4 } \left( e ^ { 2 } - 3 \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\).
4. Express \(\mathrm { g } ( x )\) in terms of \(x\). Sketch the curve \(y = \mathrm { g } ( x )\) on the copy of Fig. 8, indicating the coordinates of its intercepts with the \(x\) - and \(y\)-axes and of its turning point.
5. Write down the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\) and the \(x\)-and \(y\)-axes.

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 .

8 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

8 The equation of a curve is \(x ^ { 3 } + y ^ { 3 } = 6 x y\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Show that the point \(\left( 2 ^ { \frac { 4 } { 3 } } , 2 ^ { \frac { 5 } { 3 } } \right)\) lies on the curve and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) at this point.
3. The point \(( a , a )\), where \(a > 0\), lies on the curve. Find the value of \(a\) and the gradient of the curve at this point.

3 The equation of a curve is \(x y ^ { 2 } = x ^ { 2 } + 1\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\), and hence find the coordinates of the stationary points on the curve.