MATH101 Calculus Workshop Exercise 2 Solutions

W2.1. (i) We have

$$\eqalignno{RHS&=2\cosh x\sinh x\cr&=2{{1}\over{2}}(e^{x}+e^{-x}){{1}\over{2}}(e^{x}-e^{-x})\cr&={{1}\over{2}}(e^{2x}-e^{-2x})\cr&=\sinh 2x=LHS.\cr}$$

(ii) Similarly, we have

$$\eqalignno{RHS&=\cosh^{2}x+\sinh^{2}x\cr&={{1}\over{4}}(e^{x}+e^{-x})^{2}+{{1}\over{4}}(e^{x}-e^{-x})^{2}\cr&={{1}\over{4}}(e^{2x}+2+e^{-2x}+e^{2x}-2+e^{-2x})\cr&={{1}\over{2}}(e^{2x}+e^{-2x})\cr&=\cosh 2x=LHS.\cr}$$

W2.2. (i)

(ii) The Heaviside function is discontinuous at $x=0$, since there it jumps from $0$ to $1$; indeed, $h(x)\rightarrow 0$ as $x\rightarrow 0-$, whereas $h(x)\rightarrow 1$ as $x\rightarrow 0+.$ The graph of $h$ has a step at $x=0.$

The function $f$ is continuous as one can draw its graph without lifting pen from paper. In particular, the right and left halves of the graph meet at $x=0$ since $f(x)\rightarrow 0$ as $x\rightarrow 0+$ and $f(x)\rightarrow 0$ as $x\rightarrow 0-$.

(iii) The switch represented by $f$ would be more appropriate since it increases the current smoothly, whereas the switch $h$ would give a surge of current.

W2.3. Let $x=\sinh y$. Then

$$\sinh^{2}y+1=\cosh^{2}y$$

W2.4 With $\cos:[0,\pi]\rightarrow[-1,1]$ strictly decreasing, the inverse function is $\cos^{-1}:[-1,1]\rightarrow[0,\pi]$ which has domain $[-1,1]$ and codomain $[0,\pi]$.

W2.5 (i) We have $f(t)=(1+e^{-t})^{-1}$, so by the chain rule

$$f^{\prime}(t)=-(1+e^{-t})^{-2}(-e^{-t})=(1+e^{-t})^{-2}e^{-t}$$

while

$$f(t)\bigl(1-f(t)\bigr)={{1}\over{1+e^{-t}}}\Bigl(1-{{1}\over{1+e^{-t}}}\Bigr)={{e^{-t}}\over{(1+e^{-t})^{2}}};$$

hence

$$f^{\prime}(t)=f(t)\bigl(1-f(t)\bigr).$$

(ii) We consider

$$g(t)=f(t)-{{1}\over{2}}={{1}\over{1+e^{-t}}}-{{1}\over{2}}={{1-e^{-t}}\over{2(1+e^{-t})}};$$

while

$$g(-t)=f(-t)-{{1}\over{2}}={{1}\over{1+e^{t}}}-{{1}\over{2}}={{1-e^{t}}\over{2(1+e^{t})}}={{e^{-t}-1}\over{2(1+e^{-t})}};$$

so

$$g(t)=-g(-t).$$

(iii) Let $y=f(x)$, so

$$y={{1}\over{1+e^{-x}}},$$

hence

$$1+e^{-x}={{1}\over{y}}$$

$${\hbox{so}}\quad e^{-x}={{1}\over{y}}-1={{1-y}\over{y}}$$

hence

$$e^{x}={{y}\over{1-y}},$$

hence

$$x=\log{{y}\over{1-y}}.$$

W2.6 We have ${\hbox{sech}}^{-1}:(0,1]\rightarrow[0,\infty)$. Suppose that when $x={\hbox{sech}}^{-1}y$, we have $y={\hbox{sech}}\,x$, so

$$\eqalignno{1-y^{2}&=1-{\hbox{sech}}^{2}x\cr&=1-{{1}\over{\cosh^{2}x}}\cr&={{\cosh^{2}x-1}\over{\cosh^{2}x}}\cr&={{\sinh^{2}x}\over{\cosh^{2}x}}\cr}$$

so

$$\eqalignno{{{1+\sqrt{1-y^{2}}}\over{y}}&={{1+{{\sinh x}\over{\cosh x}}}\over{{{1}\over{\cosh x}}}}\cr&=\sinh x+\cosh x\cr&=e^{x},\cr}$$

hence

$$\log\Bigl({{1+\sqrt{1-y^{2}}}\over{y}}\Bigr)=\log e^{x}=x,$$

hence

$$x={\hbox{sech}}^{-1}y=\log\Bigl({{1+\sqrt{1-y^{2}}}\over{y}}\Bigr).$$

W2.7.

$$f(x)={{2x^{2}-4x-30}\over{x+1}}.$$

(i) When $x=0$, $y=-30$, so the intercept with the $y$ axis is at $(0,-30)$; also $y=0$ if $2x^{2}-4x-30=0,$ so $x^{2}-2x-15=0$, so $x=5$ or $x=-3$ and the intercepts with the $x$ axis are at $(5,0)$ and $(-3,0)$.

(ii) By dividing, we obtain

$$y={{2x^{2}-4x-30}\over{x+1}}=2x-6-{{24}\over{x+1}}.$$

As $x\rightarrow(-1)-$, we have $y\rightarrow\infty$; whereas as $x\rightarrow(-1)+$, we have $y\rightarrow-\infty$. Hence $x=-1$ is a vetical asymptote to the graph.

(iii) As $x\rightarrow\pm\infty$, $-24/(x+1)\rightarrow 0$, so the line $y=2x-6$ is close to the graph of $f(x)$. (Hence $y=2x-6$ is an asymptote to the graph.) Hence $f(x)\rightarrow\infty$ as $x\rightarrow\infty$, and $f(x)\rightarrow-\infty$ as $x\rightarrow-\infty$.

W2.8. (i) The function

$$f(x)={{x+1}\over{x^{2}-x-6}}$$

is defined, provided that the denominator is non-zero. The quadratic factors as

$$x^{2}-x-6=(x-3)(x+2)$$

so that the roots are $x=3,-2$; hence the domian of definition of $f$ is $(-\infty,\infty)\setminus\{3,-2\}$.

(ii) The intercept with the $y$ axis is at $f(0)=-1/6$, and the intercept with the $x$-axis is at $x=-1$.

Asymptotes are the $x$-axis, since

$$f(x)\rightarrow 0+\quad{\hbox{as}}\quad x\rightarrow\infty,\quad{\hbox{and}}\quad f(x)\rightarrow 0-\quad{\hbox{as}}\quad x\rightarrow-\infty,$$

and the vertical asymptotes $x=3$ and $x=-2$ since

$$x\rightarrow 3+$$