Refresher and Functions Tutorial Sheet, Sheet #1

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Problem sheet

Skill Building Questions

Problem 1.

Separate the following using the method of partial fractions.

(a) $\frac{x^2-6}{x^3-x}$

$\Rightarrow \quad $ factorise the denominator $\Rightarrow{}\ \ x^3-x = x(x+1)(x-1)$
$\Rightarrow{}\ \ \frac{x^2-6}{x^3-x} = \frac{A}{x} + \frac{B}{(x+1)} + \frac{C}{(x-1)}$
from numerators $x^2 - 6 = (A+B+C)x^2 + (C-B)x - A$
$\Rightarrow \quad A+B+C = 1,\\ \quad C-B = 0,\\ \quad A = 6$
$\Rightarrow \quad C = B = \frac{-5}{2}$
$\Rightarrow{}\ \ \boxed{ \frac{x^2-6}{x^3-x} = \frac{6}{x} - \frac{5}{2(x+1)} - \frac{5}{2(x-1)}}$








(b) $\frac{5(x-7)}{x^2+2x-35}$

$\Rightarrow{}\quad$ factorize the denominator
$\Rightarrow \quad x^2+2x-35 = (x+7)(x-5)$
$\Rightarrow \quad \frac{5(x-7)}{x^2+2x-35} = \frac{A}{x+7} + \frac{B}{x-5}$
$\Rightarrow \quad $ from numerators $5x-35 = (A+B)x - 5A + 7B$
$\Rightarrow \quad A+B = 5,\quad -5A + 7B = -35$
$\Rightarrow \quad \boxed{A = \frac{35}{6},\quad B=\frac{-5}{6}, \ \ \frac{5(x-7)}{x^2+2x-35} = \frac{35}{6(x+7)} - \frac{5}{6(x-5)} }$








(c) $\frac{x-1}{(3x+2)^2(2x+3)}$

$\Rightarrow{}\quad$ for linear and repeated factors
$\Rightarrow{}\ \ \frac{x-1}{(3x+2)^2(2x+3)} = \frac{A}{2x+3} + \frac{B}{3x+2} + \frac{C}{(3x+2)^2}$
$\Rightarrow{}\quad$from numerators $\Rightarrow{}\quad x-1 = (9A+6B)x^2 + (12A+13B+2C)x + 4A + 6B +3C$
$\Rightarrow{}\quad 9A+6B = 0,\quad 12A +13B +2C = 1,\quad 4A+6B+3C=-1$
$\Rightarrow{}\quad A = \frac{-2}{5},\quad B=\frac{3}{5},\quad C=-1$
$\Rightarrow{}\ \ \boxed{ \frac{x-1}{(3x+2)^2(2x+3)} = -\frac{2}{5(2x+3)} + \frac{3}{5(3x+2)} - \frac{1}{(3x+2)^2} }$








Problem 2.

Obtain the derivative of the function, $f(x)=\frac{x+3}{2-x}$, using the definition of derivative (The limit of rise over run or “$\lim{(RoR)}$”

(a) $f(x)= {x^2}$

$\Rightarrow{}\quad$ using the expression $$f'(x) = \lim_{\Delta x\to0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$$
$\Rightarrow{}\ \ f'(x) = \lim_{\Delta x\to0} \frac{(x+\Delta x)^2-(x^2)}{\Delta x}$
$\Rightarrow{}\ \ f'(x) = \lim_{\Delta x\to0} \frac{2 * \Delta x * x}{\Delta x}$
$= \boxed{2x}$








(b) $f(x)= \sin{x}$

$\Rightarrow{}\ \ f'(x) = \lim_{\Delta x\to0} \frac{\sin{x}\cos{\Delta x} + \cos{x}\sin{\Delta x} - \sin{x}}{\Delta x} = \boxed{\cos{x}}$








(c)$f(x)=\frac{x+3}{2-x}$

$\Rightarrow{}\ \ f'(x) = \lim_{\Delta x\to0} \frac{\frac{(x+\Delta x)+3}{2-(x+\Delta x)} - \frac{(x+3)}{(2-x)}}{\Delta x} = \lim_{\Delta x\to0} \frac{\frac{5\Delta x}{(2-x)(2-x-\Delta x)}}{\Delta x}$
$\Rightarrow{}\ \ f'(x) = \lim_{\Delta x\to0} \frac{5}{(2-x)(2-x-\Delta x)}= \boxed{ \frac{5}{(2-x)^2} }$








Problem 3.

Obtain the derivative of the following functions, using the product, chain and sum rule

(a) $f(x) = x^3 e^{2x}$

$\Rightarrow{}$ Using product rule $\Rightarrow{} \frac{d}{dx}(uv) = v\frac{du}{dx} + u\frac{dv}{dx}\\ \Rightarrow{} u = x^3,\quad v=\exp{(2x)}$
$\Rightarrow{} \frac{du}{dx} = 3x^2$
$\Rightarrow{}$ using chain rule $\frac{d}{dx}v(g(x)) = \frac{dv}{dg}\frac{dg}{dx}$ to solve $\frac{dv}{dx}$ with $g(x) = 2x $
$\Rightarrow{} \frac{dv}{dx} = 2\exp{(2x)}$
$\Rightarrow{} \boxed{\frac{df(x)}{dx} = 3x^2\exp{(2x)} + 2x^3\exp{(2x)} \Rightarrow{} \frac{df(x)}{dx} = x^2\exp{(2x)}(3 + 2x)}$








(b) $f(x) = \sin{(x^2 + 3x)}$

$\Rightarrow{}$ Using chain rule $\Rightarrow{} \frac{df(x)}{dx}= \cos{(x^2+3x)}\frac{d}{dx}(x^2+3x)$
$\Rightarrow{}$ Using sum rule $\Rightarrow{} \frac{d}{dx}(g(x)+h(x)) = \frac{dg}{dx} + \frac{dh}{dx}$
$\Rightarrow{} \frac{d}{dx}(x^2+3x) = 2x + 3 $
$\Rightarrow{} \boxed{\frac{df(x)}{dx} = \cos{(x^2+3x)}(2x + 3)}$








(c) $f(x) = \frac{x^2+2}{3x+1}$

$\Rightarrow{}$ Using product, chain and sum rule
$\Rightarrow{} f(x) = (x^2+2)(3x+1)^{-1}$
$\Rightarrow{} \frac{df(x)}{dx} = 2x(3x+1)^{-1} + (x^2+2)(-3(3x+1)^{-2})$
$\Rightarrow{} \boxed{\frac{df(x)}{dx} = \frac{3x^2+2x-6}{(3x+1)^{2}}}$








Problem 4.

Integrate the following functions, using the power rule, sum rule and integration by parts.

(a) $f(x) = 2x\sin{(x)}$

$\Rightarrow{} \int{2x\sin{(x)}}dx$
$\Rightarrow{}$ using integration by parts $\Rightarrow{} \int{udv} = uv -\int{vdu}$
$\Rightarrow{} u =2x,\quad dv=\sin{(x)dx} \Rightarrow{} du=2dx,\quad v=-\cos{(x)}$
$\Rightarrow{} \int{2x\sin{(x)}}dx = 2x(-\cos{(x)}) - \int{(-cox(x))2}dx$
$\Rightarrow{} \int{2x\sin{(x)}}dx = -2x\cos{(x)} + 2\int{\cos{(x)}}dx$
$\Rightarrow{} \boxed{\int{2x\sin{(x)}}dx = -2x\cos{(x)} + 2\sin{(x)} + c}$








(b) $f(x) = x^2 \ln{(4x)}$

$\Rightarrow{} \int{x^2 \ln{(4x)}}dx$
$\Rightarrow{}$ using integration by parts $\Rightarrow{} u =\ln{(4x)},\quad dv=x^2dx \Rightarrow{} du=\frac{1}{x}dx,\quad v=\frac{x^3}{3}$
$\Rightarrow{} \int{x^2 \ln{(4x)}}dx = \ln{(4x)}\frac{x^3}{3} - \int{\frac{x^3}{3}\frac{1}{x}}dx$
$\Rightarrow{} \int{x^2 \ln{(4x)}}dx = \frac{x^3}{3}\ln{(4x)} - \int{\frac{x^2}{3}}dx$
$\Rightarrow{} \int{x^2 \ln{(4x)}}dx = \frac{x^3}{3}\ln{(4x)} - \int{\frac{x^2}{3}}dx$
$\Rightarrow{} \int{x^2 \ln{(4x)}}dx = \frac{x^3}{3}\ln{(4x)} - \frac{x^3}{9} + c$
$\Rightarrow{} \boxed{\int{x^2 \ln{(4x)}}dx = \frac{x^3}{9}{(3\ln{(4x)} - 1) + c}}$








(c) $f(x) = x^5 e^{x^2}$

$\Rightarrow{} \int{x^5 \exp{(x^2)}}dx$
$\Rightarrow{}$ substitute $ u =x^2,\quad du=2xdx$
$\Rightarrow{} \int{x^5 \exp{(x^2)}}dx = \frac{1}{2}\int{u^2 \exp{(u)}}du$
$\Rightarrow{}$ using integration by parts
$\Rightarrow{} f =u^2,\quad dg=\exp{(u)}du \\ \Rightarrow{} df=2udu,\quad g=\exp{(u)}$
$\Rightarrow{} \frac{1}{2}\int{u^2 \exp{(u)}}du = \frac{1}{2} \big[u^2\exp{(u)} - \int{\exp{(u)}2u}du \big]$
$\Rightarrow{} \frac{1}{2}\int{u^2 \exp{(u)}}du = \frac{1}{2}u^2\exp{(u)} - \int{\exp{(u)}u}du$
$\Rightarrow{}$ using integration by parts $\Rightarrow{} f =u,\quad dg=\exp{(u)}du \Rightarrow{} df=du,\quad g=\exp{(u)}$
$\Rightarrow{} \frac{1}{2}\int{u^2 \exp{(u)}}du = \frac{1}{2}u^2\exp{(u)} - \big[ u\exp{(u)} - \int{\exp{(u)}}du \big]$
$\Rightarrow{} \frac{1}{2}\int{u^2 \exp{(u)}}du = \frac{1}{2}u^2\exp{(u)} - u\exp{(u)} + \exp{(u)} + c$
$\Rightarrow{} \frac{1}{2}\int{u^2 \exp{(u)}}du = \frac{1}{2}\exp{(u)}(u^2 - 2u + 2) + c$
$\Rightarrow{}$ substitute back for $ u =x^2$
$\Rightarrow{} \boxed{\int{x^5 \exp{(x^2)}}dx = \frac{1}{2}\exp{(x^2)}(x^4 - 2x^2 + 2) + c}$








Problem 5.

Express the following without logarithms (assume we are working in base 10 unless otherwise stated)

(a) $\log{x}=\log{P}+2\log{Q}-\log{K}-3$

$\log{x}=\log{\frac{PQ^2}{K}}-\log{1000\ \ \ }$
$\Rightarrow{}\ \ \boxed{=\frac{PQ^2}{1000K}}$








(b) $\log{R}=1+\frac{1}{3}\log{M}+3\log{S}$

$\Rightarrow{} \log{R=\log{10+\log{M^{\frac{1}{3}}+\log{S^3\ \ }}}}$
$\Rightarrow{}\ \boxed{R=10 \cdot S^3 \cdot \sqrt[3]{M}}$








(c) $\ln{P}=\frac{1}{2}\ln\left(Q+1\right)-3\ln{R}+2$

$\Rightarrow{} \ln{P=\ln{ \left(Q+1\right) }^{\frac{1}{2}}+\ln{e^2-\ln{R^3}}}$
$\Rightarrow{}\ \ \boxed{P=\frac{e^2\sqrt{Q+1}}{R^3}}$








Problem 6.

Express the following in log form, with no fractions or powers inside the log.

(a) $V=\frac{\pi{}h}{4}\left(D-h\right)\left(D+h\right)$

$\Rightarrow\log{V=\boxed{\log{\pi{}+\log{h+\log{(D-h)}}}+\log{\left(D+h\right)-\log4}}}$








(b) $P=\frac{1}{16}(2d-1)^2N\sqrt{S}$

$\Rightarrow\log{P}=\boxed{2\log{\left(2d-1\right)}+\log{N}+\frac{1}{2}\log{S}-\log{16}}$








Problem 7.

Solve the following equations using logarithm theorems.

(a) $\log{(5x-1)}=2$

${10}^2=5x-1\ \ \ \Rightarrow{}\ \ \ \boxed{x=\frac{101}{5}=20.2}$








(b) $\log_2{(x+1)}-\log_2{(x-4)}=3$

$\log_2{\frac{x+1}{x-4}}=3\ \ \ \Rightarrow{}\ \ \ 2^3=\frac{x+1}{x-4}$ $\Rightarrow{}\ \ \ x+1=8x-32\ \ \ \Rightarrow{}\ \ \ \boxed{x=\frac{33}{7}=4.71...}$








(c) $\log_6{(x+4)}+\log_6{(x-2)}=\log_6{(4x)}$ where $x>0$

$\log_6{\left((x+4).(x-2)\right)}=\log_6{4x}\ \ \Rightarrow{}\ \left(x+4\right)\left(x-2\right)=4x$
$\Rightarrow{}\ \ x^2-2x-8=0$
$\Rightarrow{}\ \ \ (x-4)(x+2)=0\ \ \ x_1=4 \text{ and } x_2=-2\ \ \ \Rightarrow{}\ \ \ \boxed{x=4}$








Problem 8.

Determine the domain and range of the functions defined below.

(a) $f(x)=\cos^2{(x)}$

$\Rightarrow{}\boxed{ \text{Domain of } f\left(x\right)\ \Rightarrow{}\ \ (-\infty{},\infty{})}$
$\Rightarrow{} \boxed{ \text{Range of f }\left(x\right)\Rightarrow{}[0,1].}$








(b) $f(x)={-x}^2+5x-6$

$\Rightarrow{}\boxed{\text{Domain of f}\left(x\right)\ \Rightarrow{}\ \ (-\infty{},\infty{})}$
$\Rightarrow{}\boxed{\text{Range of } f(x) \Rightarrow{} \text{, function has one maximum.} \Rightarrow{}\quad y\leq \frac{1}{4}}$








(c) $f(x)=-x^3+4x^2-4x\ $

$\Rightarrow{}\boxed{\text{ Domain of f}\left(x\right)\ \Rightarrow{}(-\infty{},\infty{})}$.
$\Rightarrow{}\boxed{\text{Range of f}\left(x\right)\Rightarrow{}\ \ (-\infty{},\infty{})}$








(d) $f\left(x\right)=$exp⁡$(1/x)$

$y=e^{\frac{1}{x}}\ \Rightarrow{}$ at $x=0$, the function is undefined. $\Rightarrow{} \boxed{\text{Domain of } f(x)\ \Rightarrow{}(-\infty{},0)\cup{}(0,\infty{})}$.
$y=e^{\frac{1}{x}}\ \Rightarrow{}$ at $y=1$, the function is undefined.
$\Rightarrow{} 0 < y < 1 \text{ & } y>1$
$ \Rightarrow{}\ \ \boxed{\text{Range of } f(x)\ \Rightarrow{}\ \ (0,1)\cup{}(1,\infty{})}$








(e) $f\left(x\right)=\frac{1}{e^x+1}$

$\Rightarrow{} \boxed{\text{Domain of the function} \Rightarrow{}(-\infty{},\infty{})}$.
$y=\frac{1}{e^x+1}\ \Rightarrow{}$ at $y=0\ \text{ & } y=1$, the function is undefined.
$\Rightarrow{}\boxed{\text{Range of } f(x)\ \Rightarrow{}\ \ (0,1)}$.








Problem 9.

Evaluate the following by applying change of base to base 10 (rounded to 1 decimal place) and then using these numbers with finite accuracy to calculate the final answer (also to 3 significant figures)

(a) $\log_2{15}$

$\log_{10}2\times{}\log_2{15}=\log_{10}{15}$ $\Rightarrow{} \boxed{\log_2{15}=\frac{\log_{10}{15}}{\log_{10}2}\approx\frac{1.2}{0.3}=4.00.}$








(b) $\log_{20}{17}$

$\log_{10}20\times{}\log_{20}{17}=\log_{10}{17}$ $\Rightarrow{} \boxed{\log_{20}{17}=\frac{\log_{10}{17}}{\log_{10}20}\approx\frac{1.2}{1.3}=0.923}$








(c) $\log_3{16}$

$\log_{10}3\times{}\log_3{16}=\log_{10}{16}$ $\Rightarrow{} \boxed{\log_3{16}=\frac{\log_{10}{16}}{\log_{10}3}\approx\frac{1.2}{0.5}=2.40}$








Problem 10.

Determine if the function $f(x)$ is odd, even, or neither.

(a) $f\left(x\right)=\ x^2\sin{(2x)}$

Break the function up: $x^2$ is an even function and $\text{sin}(2x)$ is an odd function. An even function multiplied by an odd function results in an odd function. $\\ \Rightarrow \boxed{\text{Odd function}}$








(b) $f\left(x\right)=3\sin{\left(x\right)}\cos{\left(4x\right)}$

Break the function up: $3$ is an even function, $\text{sin}(x)$ is an odd function and $\text{cos}(4x)$ is an even function. An even function multiplied by an odd function, multiplied by an even function results in an odd function. $\\ \Rightarrow \boxed{\text{Odd function}}$








(c) $f\left(x\right)=x^3e^3$

Break the function up: $x^3$ is an odd function and $e^3$ is an even function. An odd function multiplied by an even function results in an odd function. $\\ \Rightarrow \boxed{\text{Odd function}}$








(d) class-c

$\Rightarrow \boxed{\text{Odd function}}$








(e) class-c

$\Rightarrow \boxed{\text{Even function}}$








(f) class-c

$\Rightarrow \boxed{\text{Odd function}}$








Exam Style Questions

Problem 11.

A curve has equation $f(x) = \frac{ax^2 - 12}{4x^2 + bx - 6}$, where $a$ and $b$ are constants.

(a) Find the coordinates of the point where the curve crosses the $y$-axis

When $x=0, f(x)=\frac{-12}{-6} \\ \Rightarrow{} f(x) = 2,\ \\ \Rightarrow{} \boxed{(0,2)}$








(b) You are given that the curve has a vertical asymptote at $x=2$. Find the value of $b$ and the equation of the other vertical asymptote.

At $x=2, 4x^2 +bx - 6 = 0$
$\Rightarrow 4(2)^2 + b(2) - 6 = 0 \Rightarrow \boxed{b = - 5}$
$\Rightarrow$ Factorise $4x^2 - 5x - 6$
$\Rightarrow (4x+3)(x-2)$
$\therefore$ The other asymptote occurs when $4x+3=0 \Rightarrow \boxed{x = -\frac{3}{4}}$








(c) You are given that the curve crosses the $x$-axis when $x= \pm{\sqrt{6}}$. Find the value of $a$ and the equation of the horizontal asymptote.

When $f(x)=0, x = \pm{\sqrt{6}}$
$\therefore 0 = \frac{ax^2 - 12}{4x^2 + bx - 6}\\ a(6)-12 = 0 $
$6a = 12 \Rightarrow \boxed{a = 2}$
Horizontal asymptote occurs when $x \rightarrow{\infty}$
$f(x)$ will tend to the highest powers of x
$\Rightarrow f(x) = \frac{ax^2}{4x^2}$
Therefore, the horizontal asymptote occurs at $\boxed{y=\frac{1}{2}.}$








(d) Find the set of values for which $y \geqslant 0$.

Using the graph you sketched, you can see that $y$ is greater than or equal to zero in the regions where $\boxed{x\leqslant -\sqrt{6}, \space - \frac{3}{4} < x < 2, \space x \geqslant \sqrt{6}}$








(e) Find the domain (using the set notation)

$\boxed{ x \ \epsilon \ \mathbb{R}, x \neq -\frac{3}{4}, x \neq 2}$








(f) Sketch the function, indicating the location of the features listed above









Problem 12.

A curve has equation $y=\frac{3x^2 - 9}{x^2 + 3x - 4}$

(a) Find the equations of the two vertical asymptotes and the one horizontal asymptote of this curve.

$\boxed{x = 1 \\ x = -4 \\ y = 3}$








(b) State, with justification, how the curve approaches the horizontal asymptote for large positive and large negative values of x.

$\text{Large positive}\ x, \boxed{y \ \rightarrow{3} \text{ from below}}$
$\text{Large negative} \ x, \boxed{ \ y \rightarrow{3}\text{ from above}}$








(c) Sketch the curve









(d) Solve the inequality $\frac{3x^2 - 9}{x^2 + 3x -4} \geqslant 0. $

From the graph you have just sketched, you can see that the function is greater than or equal to zero in the regions where $\boxed{x<-4, -\sqrt{3} \leqslant x < 1, x \geqslant \sqrt{3}}$








Problem 13.

The following plot shows the function $f(x)= a\cos(b \pi x)+e^\frac{3x}{c}$, where the parameters $a, b$ and $c$ are all integers

Analyse the graph to work out the values of $a, b$ and $c$, explaining your reasoning in each case.

First, use the y intercept to find the value of $a$
$a + 1 = 4 \Rightarrow a = 3$
The value of $b$ can be found by looking at the periodicity. Since the oscillation period of the function is 1, $\cos(b \pi x) $ must equal $\cos(2\pi x) \Rightarrow b = 2$
Finally, $c$ is found by using any point on the graph. Since the curve passes approximately passes through the point $(0.5, -1.55)$:
$3\cos(\pi)+e^\frac{3}{2c} \approx -1.55$
$\frac{3}{2c} \approx \ln{1.45}$
$2c = 8 \Rightarrow c = 4$
$\boxed{a = 3 \\ b = 2 \\ c = 4}$








Problem 14.

The function $f(x)= \frac{e^{ax}}{bx+2}+\frac{x}{c}$, is shown in the graph below, where the parameters $a, b$ and $c$ are all integers

Analyse the graph to work out the values of $a, b$ and $c$, explaining your reasoning in each case.

As there is an asymptote at $x=-1$, set the denominator of $\frac{e^{ax}}{bx+2}$ to zero at this point
$b(-1)+2 = 0 \Rightarrow b = 2$
Next, as $x$ tends to $-\infty$, the function will approach $\frac{x}{c}$. By looking at the plot, the graph approaches $\frac{1}{4}x$
Therefore, $\frac{x}{c} = \frac{1}{4}x \Rightarrow c = 4$
Finally, $a$ can be found by using one point on the graph. Since the curve passes approximately passes through the point $(0.5, 4)$:
$\frac{e^{0.5a}}{1+2}+\frac{0.5}{4} \approx 4$
$e^{0.5a} \approx 19$
$a = 5$
$\boxed{a = 5 \\ b = 2 \\ c = 4}$








Extension Questions

Problem 15.

A curve has the equation $y=x^3-3x^2-9x+3$, and is odd (look up ‘parity’ if you don’t know what this means) about the point $P$. Find the coordinates of $P$ and use transformation arguments to justify that the curve is odd about $P$.

All cubics have rotational symmetry around their point of inflection.
$\frac{\partial y}{\partial x} = 3x^2-6x-9$
$\frac{\partial ^{2}y}{\partial x^{2}} = 6x-6$
$\Rightarrow\quad\ x = 1, \therefore y = -8$ and $P = (1, -8)$
To justify that it is odd around P, translate the graph to the origin then test.
Before:

This can be achieved in two steps:
Up by 8:
$\Rightarrow y = (x^3-3x^2-9x+3) + 8$
$\Rightarrow y = x^3-3x^2-9x+11$
Left by 1:
$\Rightarrow y = (x+1)^3-3(x+1)^2-9(x+1)+11$
$\Rightarrow y = x^3-12x$
After:

The definition of oddity is $-f(x) = f(-x)$
$f(-x) = (-x)^3-12(-x) = -x^3+12x = -f(x)$

$\boxed{\text{Therefore it is odd around the point P.}}$








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Revision Questions

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