Modulus function

Let \(f(x) = 7 - 2|x|\). A sequence \(u_0, u_1, u_2, \ldots\) is defined by \(u_0 = a\) and \(u_n = f(u_{n-1})\) for \(n > 0\).

1. 1. Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(x))\).
   2. Find all solutions of the equation \(f(f(x)) = x\).
   3. Find the values of \(a\) for which the sequence \(u_0, u_1, u_2, \ldots\) has period 2.
   4. Show that, if \(a = \frac{28}{5}\), then the sequence \(u_2, u_3, u_4, \ldots\) has period 2, but neither \(u_0\) or \(u_1\) is equal to either of \(u_2\) or \(u_3\).
2. 1. Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(f(x)))\).
   2. Consider the sequence \(u_0, u_1, u_2, \ldots\) in the cases \(a = 1\) and \(a = -\tfrac79\). Hence find all the solutions of the equation \(f(f(f(x))) = x\).
   3. Find a value of \(a\) such that the sequence \(u_3, u_4, u_5, \ldots\) has period 3, but where none of \(u_0, u_1\) or \(u_2\) is equal to any of \(u_3, u_4\) or \(u_5\).

1. Show that the function \(\mathrm{f}\), given by the single formula \(\mathrm{f}(x) = |x| - |x-5| + 1\), can be written without using modulus signs as \[\mathrm{f}(x) = \begin{cases} -4 & x \leqslant 0\,,\\ 2x - 4 & 0 \leqslant x \leqslant 5\,,\\ 6 & 5 \leqslant x\,.\end{cases}\] Sketch the graph with equation \(y = \mathrm{f}(x)\).
2. The function \(\mathrm{g}\) is given by: \[\mathrm{g}(x) = \begin{cases} -x & x \leqslant 0\,,\\ 3x & 0 \leqslant x \leqslant 5\,,\\ x + 10 & 5 \leqslant x\,.\end{cases}\] Use modulus signs to write \(\mathrm{g}(x)\) as a single formula.
3. Sketch the graph with equation \(y = \mathrm{h}(x)\), where \(\mathrm{h}(x) = x^2 - x - 4|x| + |x(x-5)|\).
4. The function \(\mathrm{k}\) is given by: \[\mathrm{k}(x) = \begin{cases} 10x & x \leqslant 0\,,\\ 2x^2 & 0 \leqslant x \leqslant 5\,,\\ 50 & 5 \leqslant x\,.\end{cases}\] Use modulus signs to write \(\mathrm{k}(x)\) as a single formula, explicitly verifying that your formula is correct.

1. The function \(\f\) is defined by \(\f(x)= |x-a| + |x-b| \), where \(a < b\). Sketch the graph of \(\f(x)\), giving the gradient in each of the regions \(x < a\), \(a < x < b\) and \(x > b\). Sketch on the same diagram the graph of \(\g(x)\), where \(\g(x)= |2x-a-b|\). What shape is the quadrilateral with vertices \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\)?
2. Show graphically that the equation \[ |x-a| + |x-b| = |x-c|\,, \] where \(a < b\), has \(0\), \(1\) or \(2\) solutions, stating the relationship of \(c\) to \(a\) and \(b\) in each case.
3. For the equation \[ |x-a| + |x-b| = |x-c|+|x-d|\,, \] where \(a < b\), \(c < d\) and \(d-c < b-a\), determine the number of solutions in the various cases that arise, stating the relationship between \(a\), \(b\), \(c\) and \(d\) in each case.

Sketch the following subsets of the \(x\)-\(y\) plane:

1. \(|x|+|y|\le 1\) ;
2. \(|x-1|+|y-1|\le 1 \) ;
3. \(|x-1|-|y+1|\le 1 \) ;
4. \(|x|\, |y-2|\le 1\) .

Find all the solutions of the equation \[|x+1|-|x|+3|x-1|-2|x-2|=x+2.\]

The function \(\mathrm{f}\) is defined for \(x<2\) by \[ \mathrm{f}(x)=2| x^{2}-x|+|x^{2}-1|-2|x^{2}+x|. \] Find the maximum and minimum points and the points of inflection of the graph of \(\mathrm{f}\) and sketch this graph. Is \(\mathrm{f}\) continuous everywhere? Is \(\mathrm{f}\) differentiable everywhere? Find the inverse of the function \(\mathrm{f}\), i.e. expressions for \(\mathrm{f}^{-1}(x),\) defined in the various appropriate intervals.

Show Solution

\[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).

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\(f\) is continuous everywhere as a sum of continuous functions. \(f\) is not differentiable at \(x = -1, 1\) Suppose \begin{align*} &&y &=x^2-4x-1 \\ &&&= (x-2)^2 -5 \\ \Rightarrow &&x &= 2\pm \sqrt{y+5} \\ \\ && y &= 3x^2+1 \\ \Rightarrow && x &= \pm \sqrt{\frac{y-1}{3}} \\ \\ && y &= -5x^2+1 \\ \Rightarrow && x &=\pm \sqrt{\frac{1-y}{5}} \\ \\ \Rightarrow && f^{-1}(y) &= \begin{cases} 2 - \sqrt{y+5} & y > 4 \\ -\sqrt{\frac{y-1}{3}} & 1 < y < 4 \\ \sqrt{\frac{1-y}{5}} & -4 < y < 1 \\ 2 + \sqrt{y+5} & y < -4 \end{cases} \end{align*}