Q1
Answer
Show that $$\sqrt[3]{x}-\cos x - 1 = 0$$ has a root in the interval $[1.4,1.5]$ and find the root correct to 3 decimal places.
$1.441$
Q2
Answer
The function $$\mathrm{f}(x) = x(x+2)(x-2)$$ has three roots. Show on a suitable diagram which of the three roots will be obtained when using the Newton-Raphson method with $x_0 = 1.5$, and state two values of $x_0$ which will cause the Newton-Raphson procedure to fail.
The $x=2$ root will be obtained. <br> Newton-Raphson fails when $3x^2-4=0$, or $x = \pm\dfrac{2\sqrt{3}}{3}$
Q3
Answer
Show that $$\mathrm{f}(x) = 2 + \tan x$$ changes sign in the interval $[1.5,1.6]$, and state, with a reason, whether $\mathrm{f}(x)$ has a root in this interval.
No root because the change of sign is caused by a discontinuity.
Q4
Answer
A student observes that there is a change of sign on the interval $[-0.5,1]$ for $$\mathrm{f}(x) = \dfrac{1}{x^3+x^2} - 1$$ The student writes:
There are no roots in this interval because there is a discontinuity in the interval.
Has the student made the correct conclusion?
No - there could be (and there is) both a root and a discontinuity in an interval.
Q5
Answer
Given $\mathrm{f}(x) = x^2-x-1$,
  1. Show that $\mathrm{f}(x) = 0$ can be rewritten in the form $x = \sqrt{x+1}$.
  2. Use the iterative formula $x_{n+1} = \sqrt{x_n+1}$ with $x_0 = 0.5$ to find an approximation to the solution to $\mathrm{f}(x)=0$ to 2 decimal places, and draw a staircase diagram to illustrate this process.
  1. Some algebraic manipulation - see video
  2. $x_1 = 1.2247...$, $x_2 = 1.4915...$, $x_3 = 1.5784...$, $x_4 = 1.6057...$, $x_5 = 1.6142...$, $x_6 = 1.6168...$, $x_7 = 1.6176... \quad x = 1.62$ to 2 d.p.
Q6
Answer
Given $\mathrm{f}(x) = x^2-x-1$,
  1. Show that $\mathrm{f}(x) = 0$ can be rewritten in the form $x = \dfrac{1}{x-1}$.
  2. Use the iterative formula $x_{n+1} = \dfrac{1}{x_n-1}$ with $x_0=-2$ to find $x_1$, $x_2$, and $x_3$, and draw a cobweb diagram to illustrate this process.
  1. Some algebraic manipulation - see video
  2. $x_1 = -0.3333...$, $x_2 = -0.75$, $x_3 = -0.5714...$
Q7
Answer
Given $$\mathrm{f}(x) = x^2-x-1$$ show, on a suitable diagram, that the iterative formula $x_{n+1} = x^2_n - 1$ does not give a solution to the equation $\mathrm{f}(x) = 0$ when $x_0 = 2$, and explain why this is the case.
Cobweb/staircase diagram does not converge. This is because the gradient of the iterative formula is not between $-1$ and $1$ around the root.
Q8
Answer
Use $x_0 = 1.2$ and the Newnton-Raphson method to find a second approximation to the root of the equation $$1 - e^x + 3\sin 2x = 0$$ and show that this answer is correct to 2 decimal places.
$x_1 = 1.162072675$ <br> Positive at $1.155$, negative at $1.165$ so this is correct to 2 d.p.
Q9
Answer
Show that there is a root of the equation $$\cos x - x^2 = 0$$ in the interval $[0.8,0.9]$, and find the root to 3 decimal places using an appropriate iterative formula.
At $x = 0.8$, greater than zero, and at $x = 0.9$, less than zero. Change of sign so there is a root. <br> Use $x_{n+1} = \sqrt{x_n}$ to get $x = 0.824$
Q10
Answer
Use the Newton-Raphson method twice, starting with $x_1 = 0.5$, to find, correct to 4 decimal places, an approximation for a root of the equation $$x^3 + 10x - 4 = 0$$
$x_2 = 0.395348837$, $x_3 = 0.393889116$, $x = 0.3939$
Q11
Answer
For $\sqrt{x}-\frac{2}{x} = 0$,
  1. Show that the solution is $1.587$ correct to 3 decimal places.
  2. Write down the exact value of the solution.
  1. $\sqrt{x}-\dfrac{2}{x} < 0$ at $x = 1.5865$, and $> 0$ at $x = 1.5875$
  2. $2^{\frac{2}{3}}$
Q12
Answer
Given $\mathrm{f}(x) = x^4 - 21x - 18$,
  1. Show that $-0.834$ is a solution to the equation $\mathrm{f}(x) = 0$ correct to 3 decimal places.
  2. Find the $x$-coordinates of any stationary points of $y=\mathrm{f}(x)$, correct to 2 decimal places.
  1. $\mathrm{f}(-0.8335) < 0$ and $\mathrm{f}(-0.8345) > 0$
  2. $1.74$
Q13
Answer
Given $\mathrm{f}(x) = e^{x-1} + 2x - 15$,
  1. Show that the equation $\mathrm{f}(x) = 0$ can be written as $x = \ln(15-2x) + 1$.
  2. Use this as the basis of an iterative formula, with $x_0 = 3$, to find $x_1$ and $x_2$ to 4 decimal places.
  3. Choose a suitable interval to show that the root of the equation $\mathrm{f}(x) = 0$ is $3.16$ to 2 decimal places.
  1. Algebra - see video.
  2. $x_1 = 3.1972$, $x_2 = 3.1524$
  3. Interval: $3.155$ (negative) and $3.165$ (positive)
Q14
Answer
Given $\mathrm{f}(x) = x^2-6x+1$,
  1. Show that the equation $\mathrm{f}(x) = 0$ can be written as $x=\dfrac{x^2+1}{6}$
  2. Use a diagram to explain why $x_0 = 2$ gives a root of the equation $\mathrm{f}(x) = 0$ using the iterative formula $x_{n+1} = \dfrac{x^2_n+1}{6}$
  3. $\mathrm{f}(x) = 0$ has a second solution at $x \approx 6$. Explain why this iterative formula can never find the second solution.
  1. Algebra - see video.
  2. A graph of $y = x$ and $y = \dfrac{x^2+1}{6}$ gives a staircase diagram to converge at a root.
  3. The gradient of the iteration formula is too high (greater than 1) around this root.
Q15
Answer
Given $\mathrm{f}(x) = 4\cot x - 8x + 3$,
  1. Show that there is a root, $\alpha$, of $\mathrm{f}(x) = 0$ in the interval $[0.8,0.9]$.
  2. Show that the equation $\mathrm{f}(x) = 0$ can be written in the form $x = \dfrac{\cos x}{2\sin x} + \dfrac{3}{8}$.
  3. Use this as the basis of an iterative formula with $x_0 = 0.85$ to calculate $x_1$ and $x_2$
  4. By considering the change of sign of $\mathrm{f}(x)$ in a suitable interval, show that $\alpha =0.831$ to 3 decimal places.
  1. $\mathrm{f}(0.8) > 0$ and $\mathrm{f}(0.9) < 0$. Change of sign so root in the interval.
  2. Algebra - see video.
  3. $x_1 = 0.814238892$, $x_2 = 0.846960185$
  4. $\mathrm{f}(0.8305) > 0$ and $\mathrm{f}(0.8315) < 0$
Q16
Answer
For the equation $\sin x - \cos 2x - 0.4 = 0$,
  1. Show that there is a root in the interval $[0.5,1]$.
  2. Use the iterative formula $x_{n+1}=\dfrac{\mathrm{arccos}(\sin x_n - 0.4)}{2}$ with $x_0 = 0.6$, writing down all the digits on your calculator, to find $x_1$, $x_2$ and $x_3$.
  1. Negative at $x=0.5$, positive at $x=1$. There is a sign change, so there is a root.
  2. $x_1 = 0.7027003998$ <br> $x_2 = 0.6609777193$ <br> $x_3 = 0.6709301678$
Q17
Answer
Given $\mathrm{f}(x) = x^3-2x^2-5x-4$,
  1. The graph of $y=\mathrm{f}(x)$ has a stationary point with $x$ coordinate $p$. Explain why $x_0=p$ is not suitable to use when applying the Newton-Raphson method.
  2. Use $x_0=3.5$ and apply the Newton-Raphson procedure twice to find an approximation to a root of the equation $\mathrm{f}(x)=0$.
  3. Show that this answer is accurate to 3 decimal places.
  1. Gradient is zero so Newton-Raphson fails.
  2. $x_1 = 3.676$, $x_2 = 3.663$.
  3. $\mathrm{f}(3.6625) < 0$ and $\mathrm{f}(3.6635) > 0$
Q18
Answer
Given $\mathrm{f}(x) = x^2-\dfrac{3}{x^2}$,
  1. Show that the equation $\mathrm{f}(x)=0$ has a root $\alpha$ in the interval $[1.3,1.4]$
  2. Use 1.3 as a first approximation to $\alpha$ and apply the Newton-Raphson procedure once to find a second approximation to $\alpha$, giving your answer to 3 decimal places.
  3. Determine if this answer is correct to 3 decimal places.
  1. $\mathrm{f}(1.3) < 0$ and $\mathrm{f}(1.4) > 0$. Change of sign so root in the interval.
  2. $x_1 = 1.316$
  3. $\mathrm{f}(1.3155) < 0$ and $\mathrm{f}(1.3165) > 0$, accurate to 3 d.p.
Q19
Answer
The equation $2x^3+x^2-1=0$ has exactly one real root.
  1. Show that the Newton Raphson formula can be written $$x_{n+1} = \dfrac{4x_n^3+x_n^2+1}{6x_n^2+2x_n}$$
  2. Use $x_1=1$ to find $x_2$ and $x_3$.
  3. Explain why $x_1=0$ cannot be used with this formula.
  1. Algebra - see video.
  2. $x_2 = 0.75$, $x_3 = \frac{2}{3}$
  3. It is a stationary point, the denominator of the Newton-Raphson formula equals zero.
Q20
Answer
The equation $$\mathrm{arcsin} (2x) - 0.5x - 0.7 = 0$$ can be rearranged to give the iterative form $$x_{n+1} = a\sin(bx_n+c)$$
  1. Find the values of the constants $a$, $b$ and $c$.
  2. Use this iterative formula with $x_0 = 0.4$ to find the solution to the equation correct to 3 decimal places.
  1. $x = \dfrac{1}{2}\sin(0.5x+0.7)$
  2. $x = 0.390$
Q21
Answer
The curve with equation $y = 2^x$ intersects the straight line with equation $y = 3 - 2x$ at the point $P$, whose $x$ coordinate is $\alpha$. Starting with $x = 0.5$, use the Newton-Raphson method to find $\alpha$ correct to 3 decimal places
$\mathrm{f}(x) = 2^x + 2x - 3$, $\mathrm{f}'(x) = 2^x\ln2 + 2$, $x = 0.692$
Q22
Answer
For the equation $x^3 - 5x^2 + 12x - 6 = 0$,
  1. Show that $\alpha = 0.6556$, to 4 decimal places, is a root.
  2. A student attempts to find $\alpha$ by using the iteration $x_{n+1} = \sqrt[3]{5x_n^3-12x_n+6}$. Explain why the student will fail to find $\alpha$.
  3. Use $x_{n+1} = \dfrac{6+5x_n^2-x_n^3}{12}$ to find $x_1$, $x_2$ and $x_3$, starting from $x_0 = 0.6$.
  1. Negative at $x = 0.65555$, positive at $x = 0.65565$. There is a sign change, so there is a root.
  2. The gradient must be between $-1$ and $1$ around the root. At $x = 0.6556$ the gradient is $-3.11$.
  3. $x_1 = 0.632$, $x_2 = 0.645390336$, $x_3 = 0.651151653$
Q23
Answer
$$\mathrm{f}(x) = 4\csc x - 4x + 1$$
  1. Show that there is a root $\alpha$ of $\mathrm{f}(x) = 0$ in the interval $[1.2,1.3]$.
  2. Show that the equation $\mathrm{f}(x) = 0$ can be written as $x = \dfrac{1}{\sin x} + \dfrac{1}{4}$.
  3. Use this as the basis of an iterative formula, with $x_0 = 1.25$, to find the values of $x_1$, $x_2$ and $x_3$ to four decimal places.
  4. By considering the change of sign in a suitable interval, verify that $\alpha = 1.291$ is correct to three decimal places.
  1. $\mathrm{f}(1.2) > 0$, $\mathrm{f}(1.3) < 0$. There is a sign change, so there is a root.
  2. Algebra - see video
  3. $x_1 = 1.3038$, $x_2 = 1.2867$, $x_3 = 1.2917$
  4. $\mathrm{f}(1.2915) < 0$ and $\mathrm{f}(1.2905) > 0$
Q24
Answer
Use the Newton-Raphson method once, starting with $0.9$ as the first approximation, to find a second approximation for a root of the equation $\dfrac{4x-3}{x^2+1}-\mathrm{e}^{-x} = 0$.
$x_1 = 0.93$. Negative at $0.925$, positive at $0.935$ so this is accurate to 2 d.p.
Q25
Answer
A curve with equation $x^3 + y = xy$ meets the line with equation $y + 3x + 1 = 0$ at the point $P$.
  1. Use an iterative method based on the formula $x_{n+1} = \dfrac{1}{2}(Ax_n^3 + Bx_n^2 + C)$, where $A$, $B$, and $C$ are constants to be found, and $x_1 = -0.35$, to find the $x$ coordinate of $P$ to 3 decimal places.
  2. Use the Newton-Raphson method with the same starting value to find the $x$ coordinate of $P$ and explain the advantage of the Newton-Raphson method.
  1. $x_{n+1} = \frac{1}{2}(x_n^3+3x_n^2-1)$. Use this many times to get $x = -0.343$
  2. $x = 0.343$, much quicker to obtain.
Q26
Answer
The curve $C$ has equation $$(x-4)^2 + (y-1)^2 = 9$$ A student wants to find where the curve crosses the $x$ axis by rearranging it into $$y = 1 + \sqrt{9-(x-4)^2}$$ The student finds that there are no roots and concludes that $C$ does not cross the $x$ axis. Is the student correct?
This equation represents the top half of a circle, so there may still be a root in the bottom half of the circle.
Q27
Answer
A curve has equation $$(x-3)^2 + (y-2)^2 = 4$$
  1. Show that the equation of the curve can be written as $y = 2 \pm \sqrt{6x-x^2-5}$
  2. A student uses the sign change method to estimate the solution to the equation $2 - \sqrt{6x-x^2-5} = 0$, and determines that there is no solution in the interval $[2.9,3.1]$. Explain whether or not the student is correct.
  1. Algebra - see video.
  2. The curve touches the $x$-axis at $x=3$ - there is a root.
Q28
Answer
A student claims there are no roots of the equation $$\sin(\sqrt{x}) + \sin(x^2) = 0$$ in the interval $[2.1,2.2]$ by using the sign change method. By finding the roots of this equation in this interval, explain why the sign change method failed in this instance.
Two roots at $2.15$ $2.19$. The student used an interval that was too large and contained multiple roots.
Q29
Answer
The curve $x^3 + xy + y^3 = 10$ meets the line $y = x+2$ at the point $A$. Use the Newton-Raphson method, starting at $x = 0.1$, to find the $x$ coordinate correct to 3 decimal places.
$x^3 + x(x+2) + (x+2)^3 = 10$. <br> $\mathrm{f}(x) = 2x^3 + 7x^2 + 14x - 2$ <br> $\mathrm{f}'(x) = 6x^2 + 14x + 14$ <br> $x = 0.134$
Q30
Answer
The curve $$y = \sin(2x) + \ln x$$ has a stationary point around $x = 1.05$. Use the Newton-Raphson method to find an approximation for the $x$ coordinate of the stationary point correct to 3 decimal places.
$2\cos(2x) + \dfrac{1}{x} = 0$. Use Newton-Raphson to get $x = 1.037$