1.07i Differentiate x^n: for rational n and sums

9. Given that \(k\) is a negative constant and that the function \(\mathrm { f } ( x )\) is defined by $$f ( x ) = 2 - \frac { ( x - 5 k ) ( x - k ) } { x ^ { 2 } - 3 k x + 2 k ^ { 2 } } , \quad x \geqslant 0$$

1. show that \(\mathrm { f } ( x ) = \frac { x + k } { x - 2 k }\)
2. Hence find \(\mathrm { f } ^ { \prime } ( x )\), giving your answer in its simplest form.
3. State, with a reason, whether \(\mathrm { f } ( x )\) is an increasing or a decreasing function. Justify your answer.

4. $$\mathrm { f } ( x ) = x + \frac { 3 } { x - 1 } - \frac { 12 } { x ^ { 2 } + 2 x - 3 } , x \in \mathbb { R } , x > 1$$

1. Show that \(\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 x + 3 } { x + 3 }\).
2. Solve the equation \(\mathrm { f } ^ { \prime } ( x ) = \frac { 22 } { 25 }\).

1. The volume \(V \mathrm {~cm} ^ { 3 }\) of a spherical balloon with radius \(r \mathrm {~cm}\) is given by the formula

$$V = \frac { 4 } { 3 } \pi r ^ { 3 }$$

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) giving your answer in simplest form. At time \(t\) seconds, the volume of the balloon is increasing according to the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 900 } { ( 2 t + 3 ) ^ { 2 } } \quad t \geqslant 0$$ Given that \(V = 0\) when \(t = 0\)
2. 1. solve this differential equation to show that $$V = \frac { 300 t } { 2 t + 3 }$$
   2. Hence find the upper limit to the volume of the balloon.
3. Find the radius of the balloon at \(t = 3\), giving your answer in cm to 3 significant figures.
4. Find the rate of increase of the radius of the balloon at \(t = 3\), giving your answer to 2 significant figures. Show your working and state the units of your answer.

1. Use proof by contradiction to prove that the curve with equation

$$y = 2 x + x ^ { 3 } + \cos x$$ has no stationary points.

3. \begin{figure}[h] \end{figure} A hollow hemispherical bowl is shown in Figure 1. Water is flowing into the bowl. When the depth of the water is \(h \mathrm {~m}\), the volume \(V \mathrm {~m} ^ { 3 }\) is given by $$V = \frac { 1 } { 12 } \pi \cdot h ^ { 2 } ( 3 - 4 h ) , \quad 0 \leqslant h \leqslant 0.25$$

1. Find, in terms of \(\pi , \frac { \mathrm { d } V } { \mathrm {~d} h }\) when \(h = 0.1\) Water flows into the bowl at a rate of \(\frac { \pi } { 800 } \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Find the rate of change of \(h\), in \(\mathrm { m } \mathrm { s } ^ { - 1 }\), when \(h = 0.1\)

2. $$f ( x ) = x ^ { 3 } - 3 x ^ { 2 } + \frac { 1 } { 2 \sqrt { x ^ { 5 } } } + 2 , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 2,3 ]\).
2. Taking 3 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\). Give your answer to 3 decimal places.

2. $$\mathrm { f } ( x ) = x ^ { 2 } - \frac { 3 } { \sqrt { x } } - \frac { 4 } { 3 x ^ { 2 } } , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [1.6,1.7]
2. Taking 1.6 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\). Give your answer to 3 decimal places.

6. $$f ( x ) = x ^ { 3 } - \frac { 1 } { 2 x } + x ^ { \frac { 3 } { 2 } } , \quad x > 0$$ The root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval [0.6, 0.7].

1. Taking 0.6 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.
2. Show that your answer to part (a) is correct to 3 decimal places.

1. $$f ( x ) = 3 x ^ { 2 } - \frac { 5 } { 3 \sqrt { x } } - 6 , \quad x > 0$$ The single root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval [1.5, 1.6].

1. Taking 1.5 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.
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2. Use linear interpolation once on the interval [1.5, 1.6] to find another approximation to \(\alpha\). Give your answer to 3 decimal places.

4. $$f ( x ) = 1 - \frac { 1 } { 8 x ^ { 4 } } + \frac { 2 } { 7 \sqrt { x ^ { 7 } } } \quad x > 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a single root, \(\alpha\), that lies in the interval \([ 0.15,0.25 ]\)

1. 1. Determine \(\mathrm { f } ^ { \prime } ( x )\)
   2. Explain why 0.25 cannot be used as an initial approximation for \(\alpha\) in the Newton-Raphson process.
   3. Taking 0.15 as a first approximation to \(\alpha\) apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\) Give your answer to 3 decimal places.
2. Use linear interpolation once on the interval \([ 0.15,0.25 ]\) to find another approximation to \(\alpha\) Give your answer to 3 decimal places.

$$f ( x ) = x - 4 - \cos ( 5 \sqrt { x } ) \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [2.5, 3.5]
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2. Use linear interpolation once on the interval [2.5, 3.5] to find an approximation to \(\alpha\), giving your answer to 2 decimal places.
   (ii) $$\operatorname { g } ( x ) = \frac { 1 } { 10 } x ^ { 2 } - \frac { 1 } { 2 x ^ { 2 } } + x - 11 \quad x > 0$$
3. Determine \(\mathrm { g } ^ { \prime } ( x )\). The equation \(\mathrm { g } ( x ) = 0\) has a root \(\beta\) in the interval [6,7]
4. Using \(x _ { 0 } = 6\) as a first approximation to \(\beta\), apply the Newton-Raphson procedure once to \(\mathrm { g } ( x )\) to find a second approximation to \(\beta\), giving your answer to 3 decimal places.

4. $$\mathrm { f } ( x ) = x ^ { \frac { 3 } { 2 } } - 3 x ^ { \frac { 1 } { 2 } } - 3 , \quad x > 0$$ Given that \(\alpha\) is the only real root of the equation \(\mathrm { f } ( x ) = 0\),

1. show that \(4 < \alpha < 5\)
2. Taking 4.5 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places.
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3. Use linear interpolation once on the interval [4,5] to find another approximation to \(\alpha\), giving your answer to 3 decimal places.

1. $$f ( x ) = x ^ { 3 } - \frac { 10 \sqrt { x } - 4 x } { x ^ { 2 } } \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval[1.4,1.5]
2. Determine \(\mathrm { f } ^ { \prime } ( x )\) .
3. Using \(x _ { 0 } = 1.4\) as a first approximation to \(\alpha\) ,apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to calculate a second approximation to \(\alpha\) ,giving your answer to 3 decimal places. \(f ( x ) = x ^ { 3 } - \frac { 10 \sqrt { x } - 4 x } { x ^ { 2 } } \quad x > 0\)
   1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval[1.4,1.5]
   2. Determine \(\mathrm { f } ^ { \prime } ( x )\) .

2. $$f ( x ) = 10 - 2 x - \frac { 1 } { 2 \sqrt { x } } - \frac { 1 } { x ^ { 3 } } \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [0.4, 0.5]
2. Determine \(\mathrm { f } ^ { \prime } ( x )\).
3. Using \(x _ { 0 } = 0.5\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places. The equation \(\mathrm { f } ( x ) = 0\) has another root \(\beta\) in the interval [4.8, 4.9]
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4. Use linear interpolation once on the interval [4.8, 4.9] to find an approximation to \(\beta\), giving your answer to 3 decimal places.

7. $$f ( x ) = x ^ { \frac { 3 } { 2 } } + x - 3$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval \([ 1,2 ]\) [0pt]
2. Starting with the interval [1, 2], use interval bisection twice to show that \(\alpha\) lies in the interval [1.25, 1.5]
3. 1. Determine \(\mathrm { f } ^ { \prime } ( x )\)
   2. Using 1.375 as a first approximation for \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to determine a second approximation for \(\alpha\), giving your answer to 3 decimal places.
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4. Use linear interpolation once on the interval [1.25,1.5] to obtain a different approximation for \(\alpha\), giving your answer to 3 decimal places.

3. $$\mathrm { f } ( x ) = x ^ { 3 } - 5 \sqrt { x } - 4 x + 7 \quad x \geqslant 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 0.25,1 ]\)

1. Use linear interpolation once on the interval [ \(0.25,1\) ] to determine an approximation to \(\alpha\), giving your answer to 3 decimal places. The equation \(\mathrm { f } ( x ) = 0\) has another root \(\beta\) in the interval [1.5, 2.5]
2. Determine \(\mathrm { f } ^ { \prime } ( x )\)
3. Hence, using \(x _ { 0 } = 1.75\) as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to determine a second approximation to \(\beta\), giving your answer to 3 decimal places.

2. $$f ( x ) = 7 \sqrt { x } - \frac { 1 } { 2 } x ^ { 3 } - \frac { 5 } { 3 x } \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval [2.8, 2.9]
2. 1. Find \(\mathrm { f } ^ { \prime } ( x )\).
   2. Hence, using \(x _ { 0 } = 2.8\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to calculate a second approximation to \(\alpha\), giving your answer to 3 decimal places.
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3. Use linear interpolation once on the interval [2.8, 2.9] to find another approximation to \(\alpha\). Give your answer to 3 decimal places.

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3. $$\mathrm { f } ( x ) = x ^ { 2 } + \frac { 3 } { x } - 1 , \quad x < 0$$ The only real root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval \([ - 2 , - 1 ]\).

1. Taking - 1.5 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 2 decimal places.
2. Show that your answer to part (a) gives \(\alpha\) correct to 2 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{38217fcb-8f26-49ac-9bb1-61c2f304006e-06_2250_51_317_1980}

2. $$f ( x ) = 5 x ^ { 2 } - 4 x ^ { \frac { 3 } { 2 } } - 6 , \quad x \geqslant 0$$ The root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval \([ 1.6,1.8 ]\)

1. Use linear interpolation once on the interval \([ 1.6,1.8 ]\) to find an approximation to \(\alpha\). Give your answer to 3 decimal places.
2. Taking 1.7 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.

5. $$f ( x ) = 3 \sqrt { } x + \frac { 18 } { \sqrt { } x } - 20$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [1.1,1.2].
2. Find \(\mathrm { f } ^ { \prime } ( x )\).
3. Using \(x _ { 0 } = 1.1\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 significant figures.

3. $$f ( x ) = 5 x ^ { 2 } - 4 x ^ { \frac { 3 } { 2 } } - 6 , \quad x \geqslant 0$$ The root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval \([ 1.6,1.8 ]\).

1. Use linear interpolation once on the interval \([ 1.6,1.8 ]\) to find an approximation to \(\alpha\). Give your answer to 3 decimal places.
2. Differentiate \(\mathrm { f } ( x )\) to find \(\mathrm { f } ^ { \prime } ( x )\).
3. Taking 1.7 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.

3. $$\mathrm { f } ( x ) = 2 x ^ { \frac { 1 } { 2 } } + x ^ { - \frac { 1 } { 2 } } - 5 , \quad x > 0$$

1. Find \(\mathrm { f } ^ { \prime } ( x )\). The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [4.5, 5.5].
2. Using \(x _ { 0 } = 5\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 significant figures.

4. $$f ( x ) = 2 x ^ { \frac { 1 } { 2 } } - \frac { 6 } { x ^ { 2 } } - 3 , \quad x > 0$$ A root \(\beta\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval \([ 3,4 ]\).
Taking 3.5 as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\beta\). Give your answer to 3 decimal places.

4. $$f ( x ) = x ^ { 2 } + \frac { 5 } { 2 x } - 3 x - 1 , \quad x \neq 0$$

1. Use differentiation to find \(\mathrm { f } ^ { \prime } ( x )\). The root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval [0.7, 0.9].
2. Taking 0.8 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.

2. $$f ( x ) = 3 x ^ { \frac { 3 } { 2 } } - 25 x ^ { - \frac { 1 } { 2 } } - 125 , \quad x > 0$$

1. Find \(f ^ { \prime } ( x )\). The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [12, 13].
2. Using \(x _ { 0 } = 12.5\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places.

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