1.07n Stationary points: find maxima, minima using derivatives

8 A curve has equation \(y = ( x + 2 ) \mathrm { e } ^ { - x }\).

1. Find the coordinates of the points where the curve cuts the axes.
2. Find the coordinates of the stationary point, S , on the curve.
3. By evaluating \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at S , determine whether the stationary point is a maximum or a minimum.
4. Sketch the curve in the domain \(- 3 < x < 3\).
5. Find where the normal to the curve at the point \(( 0,2 )\) cuts the curve again.
6. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).

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 \(\mathrm { OA } ^ { \prime } = \mathrm { A } ^ { \prime } \mathrm { B } ^ { \prime }\).

8 Fig. 8 shows part of the graph of the function \(y = 5 x ( 2 x - 1 ) ^ { 3 }\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of S , the turning point of the curve.
2. Find the area of the shaded region enclosed between the curve and the \(x\)-axis.
3. Given that \(\mathrm { f } ( x ) = 5 x ( 2 x - 1 ) ^ { 3 }\), show that \(\mathrm { f } ( x + 0.5 ) = 40 x ^ { 3 } ( x + 0.5 )\).
4. Find \(\int _ { - \frac { 1 } { 2 } } ^ { 0 } 40 x ^ { 3 } ( x + 0.5 ) \mathrm { d } x\).
5. Explain, with the aid of a sketch, the connection between your answer to parts (ii) and (iv).

2 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.

2 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), which has a \(y\)-intercept at \(\mathrm { P } ( 0,3 )\), a minimum point at \(\mathrm { Q } ( 1,2 )\), and an asymptote \(x = - 1\). \begin{figure}[h] \end{figure}

1. Find the coordinates of the images of the points P and Q when the curve \(y = \mathrm { f } ( x )\) is transformed to
   (A) \(y = 2 \mathrm { f } ( x )\),
   (B) \(y = \mathrm { f } ( x + 1 ) + 2\). You are now given that \(\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { x + 1 } , x \neq - 1\).
2. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence find the coordinates of the other turning point on the curve \(y = \mathrm { f } ( x )\).
3. Show that \(\mathrm { f } ( x - 1 ) = x - 2 + \frac { 4 } { x }\).
4. Find \(\int _ { a } ^ { b } \left( x - 2 + \frac { 4 } { x } \right) \mathrm { d } x\) in terms of \(a\) and \(b\). Hence, by choosing suitable values for \(a\) and \(b\), find the exact area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\).

2 A curve has equation \(x ^ { 2 } + 2 y ^ { 2 } = 4 x\).

1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
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2. Hence find the exact coordinates of the stationary points of the curve. [You need not determine their nature.]

4 Fig. 7 shows the curve \(x ^ { 3 } + y ^ { 3 } = 3 x y\). The point P is a turning point of the curve. \begin{figure}[h] \end{figure}

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x ^ { 2 } } { y ^ { 2 } - x }\).
2. Hence find the exact \(x\)-coordinate of P .

6 A curve is defined by the equation \(( x + y ) ^ { 2 } = 4 x\). The point \(( 1,1 )\) lies on this curve.
By differentiating implicitly, show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x + y } - 1\).
Hence verify that the curve has a stationary point at \(( 1,1 )\).

2 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 \(x = \frac { 1 } { 8 }\).

3 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 \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

4 A curve has equation \(2 y ^ { 2 } + y = 9 x ^ { 2 } + 1\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point \(\mathrm { A } ( 1,2 )\).
2. Find the coordinates of the points on the curve at which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).

1 Fig. 7 shows the curve \(y = _ { x - 1 }\). It has a minimum at the point P . The line \(l\) is an asymptote to the curve. \begin{figure}[h] \end{figure}

1. Write down the equation of the asymptote \(l\).
2. Find the coordinates of P .
3. Using the substitution \(u = x - 1\), show that the area of the region enclosed by the \(x\)-axis, the curve and the lines \(x = 2\) and \(x = 3\) is given by $$\int _ { 1 } ^ { 2 } \left( u + 2 + \frac { 4 } { u } \right) \mathrm { d } u$$ Evaluate this area exactly.
4. Another curve is defined by the equation \(\mathrm { e } ^ { y } = \frac { x ^ { 2 } + 3 } { x - 1 }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\) by differentiating implicitly. Hence find the gradient of this curve at the point where \(x = 2\).

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

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

2 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 3 .
2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).

1 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.

1 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.

2 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 $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( \begin{array} { l l } u ^ { \frac { 3 } { 2 } } & u ^ { \frac { 1 } { 2 } } \end{array} \right) \mathrm { d } u .$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

3 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 \(\frac { 1 } { 6 } \ln 3\).

2

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

2 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.

2 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 \(\frac { 1 } { 6 } \ln 3\).

1 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 .

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

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } - 2 x y } { x ^ { 2 } - 2 x y }\).
2. (a) Show that, if \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\), then \(y = 2 x\).
   (b) Hence find the coordinates of the point on the curve where the tangent is parallel to the \(x\)-axis.

5

1. For the curve \(2 x ^ { 2 } + x y + y ^ { 2 } = 14\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Deduce that there are two points on the curve \(2 x ^ { 2 } + x y + y ^ { 2 } = 14\) at which the tangents are parallel to the \(x\)-axis, and find their coordinates.

6. (i) Use the derivative of \(\cos x\) to prove that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sec x ) = \sec x \tan x$$ The curve \(C\) has the equation \(y = \mathrm { e } ^ { 2 x } \sec x , - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }\).
(ii) Find an equation for the tangent to \(C\) at the point where it crosses the \(y\)-axis.
(iii) Find, to 2 decimal places, the \(x\)-coordinate of the stationary point of \(C\).