Problems

1. The continuous function \(\f\) is defined by \[ \tan \f(x) = x \ \ \ \ \ (-\infty < x <\infty) \] and \(\f(0)=\pi\). Sketch the curve \(y=\f(x)\).
2. The continuous function \(\g\) is defined by \[ \tan \g(x) = \frac x {1+x^2} \ \ \ \ \ \ (-\infty < x <\infty) \] and \(\g(0)=\pi\). Sketch the curves \(y= \dfrac x {1+x^2} \ \) and \(y=\g(x)\).
3. The continuous function \(\h \) is defined by \(\h (0)=\pi\) and \[ \tan \h (x)= \frac x {1-x^2}\, \ \ \ \ \ (x \ne \pm 1) \,. \] (The values of \(\h (x)\) at \(x=\pm1\) are such that \(\h (x)\) is continuous at these points.) Sketch the curves \(y= \dfrac x {1-x^2} \ \) and \(y=\h (x)\).
4. [Not on original exam] The continuous functions \(\h_1\) and \(\h_2\) are defined by: \(\h_1(0)=\h_2(0)=\pi \), \[ \tan \h_1(x) = \frac {x+x^4} {1+x^2+x^4} \ \ \ \ \ \text{and} \ \ \ \ \ \ \tan \h_2(x) = \frac {4x-x^3} {1-x^4} \,. \] for values of \(x\) at which the right hand sides are defined. Find \(\lim\limits_{x\to\infty}\h_1(x)\) and \(\lim\limits_{x\to\infty}\h_2(x)\,\).

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Solution:

1. \(\,\)
   

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2. \(\,\)
   

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3. \(\,\)
   

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4. Note that \(\frac{x+x^4}{1+x^2+x^4}\) is continuous, and nicely behaved on \((-\infty, \infty)\) so we can see that \(\lim_{x \to \infty} h_1(x) = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\). \(\frac{4x-x^3}{1-x^4}\) on the other hand has asymptotes at \(\pm 1\). So as as \(x \to 1\), \(h_1(x) \to \pi + \frac{\pi}{2} = \frac{3\pi}{2}\). Then as \(x \to \infty\) we increase by another \(\frac{\pi}{2}\), so \(\lim_{x \to \infty} h_2(x) = 2\pi\)

An alternative way to think about the last two parts is to consider \(h\) as giving the (continuous) argument (shifted by \(\pi\)) of \((1-t^2)+it\) (blue), \((1+t^2+t^4)+i(t+t^4)\) (orange) or \((1-t^4)+i(4t-t^3)\) (green). We can see the orange line never wraps around the origin, so the argument is always easy to find. The blue does one full circuit, from \(-\pi\) to \(\pi\) (or \(0\) to \(2\pi\) in our world. And the green line also does a full \(2\pi\) loop.

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Find \(y\) in terms of \(x\), given that: \begin{eqnarray*} \mbox{for \(x < 0\,\)}, && \frac{\d y}{\d x} = -y \mbox{ \ \ and \ \ } y = a \mbox{ when } x = -1\;; \\ \mbox{for \(x > 0\,\)}, && \frac{\d y}{\d x} = y \mbox{ \ \ \ \ and \ \ } y = b \ \mbox{ when } x = 1\;. \end{eqnarray*} Sketch a solution curve. Determine the condition on \(a\) and \(b\) for the solution curve to be continuous (that is, for there to be no `jump' in the value of \(y\)) at \(x = 0\). Solve the differential equation \[ \frac{\d y}{\d x} = \left\vert \e^x-1\right\vert y \] given that \(y=\e^{\e}\) when \(x=1\) and that \(y\) is continuous at \(x=0\,\). Write down the following limits: \ \[ \text{(i)} \ \ \lim_ {x \to +\infty} y\exp(-\e^x)\;; \ \ \ \ \ \ \ \ \ \text{(ii)} \ \ \lim_{x \to -\infty}y \e^{-x}\,. \]

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.

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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*}