1.07n Stationary points: find maxima, minima using derivatives

3 A curve has equation \(y = x + \frac { 1 } { x }\).
Use calculus to show that the curve has a turning point at \(x = 1\).
Show also that this point is a minimum.

4 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).

5 Differentiate \(4 x ^ { 2 } + \frac { 1 } { x }\) and hence find the \(x\)-coordinate of the stationary point of the curve \(y = 4 x ^ { 2 } + \frac { 1 } { x }\). \begin{figure}[h] \end{figure} The equation of the curve shown in Fig. 11 is \(y = x ^ { 3 } - 6 x + 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find, in exact form, the range of values of \(x\) for which \(x ^ { 3 } - 6 x + 2\) is a decreasing function.
3. Find the equation of the tangent to the curve at the point \(( - 1,7 )\). Find also the coordinates of the point where this tangent crosses the curve again.

1 The equation of a curve is \(y = 7 + 6 x - x ^ { 2 }\).

1. Use calculus to find the coordinates of the turning point on this curve. Find also the coordinates of the points of intersection of this curve with the axes, and sketch the curve.
2. Find \(\int _ { 1 } ^ { 5 } \left( 7 + 6 x - x ^ { 2 } \right) \mathrm { d } x\), showing your working.
3. The curve and the line \(y = 12\) intersect at \(( 1,12 )\) and \(( 5,12 )\). Using your answer to part (ii), find the area of the finite region between the curve and the line \(y = 12\).

2 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. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-2_1020_940_244_679} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} Fig. 11 shows a sketch of the cubic curve \(y = \mathrm { f } ( x )\). The values of \(x\) where it crosses the \(x\)-axis are - 5 , - 2 and 2 , and it crosses the \(y\)-axis at \(( 0 , - 20 )\).
4. Express \(\mathrm { f } ( x )\) in factorised form.
5. Show that the equation of the curve may be written as \(y = x ^ { 3 } + 5 x ^ { 2 } - 4 x - 20\).
6. Use calculus to show that, correct to 1 decimal place, the \(x\)-coordinate of the minimum point on the curve is 0.4 . Find also the coordinates of the maximum point on the curve, giving your answers correct to 1 decimal place.
7. State, correct to 1 decimal place, the coordinates of the maximum point on the curve \(y = \mathrm { f } ( 2 x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-3_768_1023_223_598} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} Fig. 11 shows the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\).
8. Use calculus to find \(\int _ { 1 } ^ { 3 } \left( x ^ { 3 } - 3 x ^ { 2 } - x + 3 \right) \mathrm { d } x\) and state what this represents.
9. Find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\), giving your answers in surd form. Hence state the set of values of \(x\) for which \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\) is a decreasing function.
10. Differentiate \(x ^ { 3 } - 3 x ^ { 2 } - 9 x\). Hence find the \(x\)-coordinates of the stationary points on the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - 9 x\), showing which is the maximum and which the minimum.
11. Find, in exact form, the coordinates of the points at which the curve crosses the \(x\)-axis.
12. Sketch the curve. A curve has equation \(y = x ^ { 3 } - 6 x ^ { 2 } + 12\).
13. Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.
14. Find, in the form \(y = m x + c\), the equation of the normal to the curve at the point \(( 2 , - 4 )\).

2 A curve has equation \(y = x ^ { 3 } - 6 x ^ { 2 } + 12\).

1. Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.
2. Find, in the form \(y = m x + c\), the equation of the normal to the curve at the point \(( 2 , - 4 )\).

3 A cubic curve has equation \(y = x ^ { 3 } - 3 x ^ { 2 } + 1\).

1. Use calculus to find the coordinates of the turning points on this curve. Determine the nature of these turning points.
2. Show that the tangent to the curve at the point where \(x = - 1\) has gradient 9 . Find the coordinates of the other point, P , on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P . Show that the area of the triangle bounded by the normal at P and the \(x\) - and \(y\)-axes is 8 square units.

3

1. A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve $$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$ where \(x\) and \(y\) are horizontal and vertical distances in metres. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_512_819_493_700} \captionsetup{labelformat=empty} \caption{Figure 9.1}
   \end{figure} Using this model,
   (A) find the greatest height of the tunnel,
   (B) explain why \(100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x\) gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
   [0pt] [5]
2. The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_506_942_1703_629} \captionsetup{labelformat=empty} \caption{Not to scale}
   \end{figure} Fig. 9.2 Use the trapezium rule with 5 strips to estimate the new cross-sectional area.
   Hence estimate the volume of earth removed when the tunnel is re-shaped.

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

7. The curve with equation \(y = x ^ { \frac { 5 } { 2 } } \ln \frac { x } { 4 } , x > 0\) crosses the \(x\)-axis at the point \(P\).

1. Write down the coordinates of \(P\). The normal to the curve at \(P\) crosses the \(y\)-axis at the point \(Q\).
2. Find the area of triangle \(O P Q\) where \(O\) is the origin. The curve has a stationary point at \(R\).
3. Find the \(x\)-coordinate of \(R\) in exact form.

1. \hspace{0pt} [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are unit vectors, with \(\mathbf { i }\) horizontal and \(\mathbf { j }\) vertical.]

\begin{figure}[h] \end{figure} The fixed points \(A\) and \(B\) lie on horizontal ground.
At time \(t = 0\), a particle \(P\) is projected from \(A\) with velocity \(( 4 \mathbf { i } + 7 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\) Particle \(P\) moves freely under gravity and hits the ground at \(B\), as shown in Figure 6 .

1. Find the distance \(A B\). The speed of \(P\) is less than \(5 \mathrm {~ms} ^ { - 1 }\) for an interval of length \(T\) seconds.
2. Find the value of \(T\) At the instant when the direction of motion of \(P\) is perpendicular to the initial direction of motion of \(P\), the particle is \(h\) metres above the ground.
3. Find the value of \(h\).

8 \includegraphics[max width=\textwidth, alt={}, center]{1216a06e-7e14-48d7-a7ca-7acd8d71af5f-4_538_1443_262_351} The diagram shows the curve with equation \(y = x ^ { 8 } \mathrm { e } ^ { - x ^ { 2 } }\). The curve has maximum points at \(P\) and \(Q\). The shaded region \(A\) is bounded by the curve, the line \(y = 0\) and the line through \(Q\) parallel to the \(y\)-axis. The shaded region \(B\) is bounded by the curve and the line \(P Q\).

1. Show by differentiation that the \(x\)-coordinate of \(Q\) is 2 .
2. Use Simpson's rule with 4 strips to find an approximation to the area of region \(A\). Give your answer correct to 3 decimal places.
3. Deduce an approximation to the area of region \(B\).

7 A curve has equation \(y = \frac { x \mathrm { e } ^ { 2 x } } { x + k }\), where \(k\) is a non-zero constant.

1. Differentiate \(x \mathrm { e } ^ { 2 x }\), and show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \mathrm { e } ^ { 2 x } \left( 2 x ^ { 2 } + 2 k x + k \right) } { ( x + k ) ^ { 2 } }\).
2. Given that the curve has exactly one stationary point, find the value of \(k\), and determine the exact coordinates of the stationary point.

6

1. Find the exact value of the \(x\)-coordinate of the stationary point of the curve \(y = x \ln x\).
2. The equation of a curve is \(y = \frac { 4 x + c } { 4 x - c }\), where \(c\) is a non-zero constant. Show by differentiation that this curve has no stationary points.

9 \includegraphics[max width=\textwidth, alt={}, center]{5c501214-b41c-43a8-b9c6-986758e83e7d-4_534_935_264_605} The function f is defined for the domain \(x \geqslant 0\) by $$f ( x ) = \frac { 15 x } { x ^ { 2 } + 5 }$$ The diagram shows the curve with equation \(y = \mathrm { f } ( x )\).

1. Find the range of f .
2. The function g is defined for the domain \(x \geqslant k\) by $$\mathrm { g } ( x ) = \frac { 15 x } { x ^ { 2 } + 5 }$$ Given that g is a one-one function, state the least possible value of \(k\).
3. Show that there is no point on the curve \(y = \mathrm { g } ( x )\) at which the gradient is - 1 .

8 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-4_476_608_287_756} The diagram shows the curve \(y = ( \ln x ) ^ { 2 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. The point \(P\) on the curve is the point at which the gradient takes its maximum value. Show that the tangent at \(P\) passes through the point \(( 0 , - 1 )\).

7 Fig. 7 shows the curve $$y = 2 x - x \ln x , \text { where } x > 0 .$$ The curve crosses the \(x\)-axis at A , and has a turning point at B . The point C on the curve has \(x\)-coordinate 1 . Lines CD and BE are drawn parallel to the \(y\)-axis. \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of A , giving your answer in terms of e .
2. Find the exact coordinates of B .
3. Show that the tangents at A and C are perpendicular to each other.
4. Using integration by parts, show that $$\int x \ln x \mathrm {~d} x = \frac { 1 } { 2 } x ^ { 2 } \ln x - \frac { 1 } { 4 } x ^ { 2 } + c$$ Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines CD and BE . \section*{[Question 8 is printed overleaf.]}

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 part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h] \end{figure}

1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

7 Fig. 7 shows the curve \(y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }\). It is undefined at \(x = a\); the line \(x = a\) is a vertical asymptote. \begin{figure}[h] \end{figure}

1. Calculate the value of \(a\), giving your answer correct to 3 significant figures.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }\). Hence determine the coordinates of the turning points of the curve.
3. Show that the area of the region between the curve and the \(x\)-axis from \(x = 0\) to \(x = 1\) is \(\frac { 1 } { 6 } \ln 3\).

5 Show that the curve \(y = x ^ { 2 } \ln x\) has a stationary point when \(x = \frac { 1 } { \sqrt { \mathrm { e } } }\).

5 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).

8 Fig. 8 shows the curve \(y = 3 \ln x + x - x ^ { 2 }\).
The curve crosses the \(x\)-axis at P and Q , and has a turning point at R . The \(x\)-coordinate of Q is approximately 2.05 . \begin{figure}[h] \end{figure}

1. Verify that the coordinates of P are \(( 1,0 )\).
2. Find the coordinates of R , giving the \(y\)-coordinate correct to 3 significant figures. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that R is a maximum point.
3. Find \(\int \ln x \mathrm {~d} x\). Hence calculate the area of the region enclosed by the curve and the \(x\)-axis between P and Q , giving your answer to 2 significant figures.

4

1. Show that \(y = a x e ^ { - x }\) for \(a > 0\) has only one stationary point for all values of \(x\). Determine whether this stationary value is a maximum or minimum point.
2. Sketch the curve.

8 You are given that \(\mathrm { f } ( x ) = \frac { x } { x ^ { 2 } + 1 }\) for all real values of \(x\).

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 1 - x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }\).
2. Hence show that there is a stationary value at \(\left( 1 , \frac { 1 } { 2 } \right)\) and find the coordinates of the other stationary point.
3. The graph of the curve is shown in Fig. 8. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-3_518_892_1612_705} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure} State whether the curve is odd or even and prove the result algebraically.
4. Show that \(\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 1 } \mathrm {~d} x = \int _ { a } ^ { b } k \frac { 1 } { u + 1 } \mathrm {~d} u\), where the values of \(a , b\) and \(k\) are to be determined.
5. Hence find the area of the shaded region in Fig. 8.