Refresher and Functions Tutorial Sheet, Sheet #1
Learning Targets
- Use the method of partial fractions
- Find the derivative of functions with the product, chain and sum rules
- Find the integral of functions using the power rule, sum rule and integration by parts
- Simplify logarithms
- Identify key points in a function (roots, asymptotes, stationary points, inflection points, domain, range)
- Sketch a graph of a given function
Additional Resources
Tutorials
- Curve Sketching : Extra video by Sam recapping curve sketching from the lecture.
- Calculus Playlist : Series with some great animations for developing intuition - visualisation and understanding is key to success in this module!
Software
- WolframAlpha : Use this to plot any function - you will become very familiar with this over the module.
Problem sheet
Skill Building Questions
Problem 1.
Separate the following using the method of partial fractions.
(a) $\frac{x^2-6}{x^3-x}$
$\Rightarrow{}\ \ \frac{x^2-6}{x^3-x} = \frac{A}{x} + \frac{B}{(x+1)} + \frac{C}{(x-1)}$
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) $\frac{5(x-7)}{x^2+2x-35}$
$\Rightarrow \quad x^2+2x-35 = (x+7)(x-5)$
$\Rightarrow \quad \frac{5(x-7)}{x^2+2x-35} = \frac{A}{x+7} + \frac{B}{x-5}$
$\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) $\frac{x-1}{(3x+2)^2(2x+3)}$
$\Rightarrow{}\ \ \frac{x-1}{(3x+2)^2(2x+3)} = \frac{A}{2x+3} + \frac{B}{3x+2} + \frac{C}{(3x+2)^2}$
$\Rightarrow{}\quad$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) $f(x)= {x^2}$
$\Rightarrow{}\ \ f'(x) = \lim_{\Delta x\to0} \frac{(x+\Delta x)^2-(x^2)}{\Delta x}$
$\Rightarrow{}\ \ f'(x) = \lim_{\Delta x\to0} \frac{2 * \Delta x * x}{\Delta x}$
$= \boxed{2x}$
(b) $f(x)= \sin{x}$
(c)$f(x)=\frac{x+3}{2-x}$
$\Rightarrow{}\ \ f'(x) = \lim_{\Delta x\to0} \frac{5}{(2-x)(2-x-\Delta x)}= \boxed{ \frac{5}{(2-x)^2} }$
Problem 3.
Obtain the derivative of the following functions, using the product, chain and sum rule
(a) $f(x) = x^3 e^{2x}$
$\Rightarrow{} \frac{du}{dx} = 3x^2$
$\Rightarrow{}$ 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 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) $f(x) = \frac{x^2+2}{3x+1}$
$\Rightarrow{} f(x) = (x^2+2)(3x+1)^{-1}$
$\Rightarrow{} \frac{df(x)}{dx} = 2x(3x+1)^{-1} + (x^2+2)(-3(3x+1)^{-2})$
$\Rightarrow{} \boxed{\frac{df(x)}{dx} = \frac{3x^2+2x-6}{(3x+1)^{2}}}$
Problem 4.
Integrate the following functions, using the power rule, sum rule and integration by parts.
(a) $f(x) = 2x\sin{(x)}$
$\Rightarrow{}$ 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) $f(x) = x^2 \ln{(4x)}$
$\Rightarrow{}$ 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) $f(x) = x^5 e^{x^2}$
$\Rightarrow{}$ 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) $\log{x}=\log{P}+2\log{Q}-\log{K}-3$
$\Rightarrow{}\ \ \boxed{=\frac{PQ^2}{1000K}}$
(b) $\log{R}=1+\frac{1}{3}\log{M}+3\log{S}$
$\Rightarrow{}\ \boxed{R=10 \cdot S^3 \cdot \sqrt[3]{M}}$
(c) $\ln{P}=\frac{1}{2}\ln\left(Q+1\right)-3\ln{R}+2$
$\Rightarrow{}\ \ \boxed{P=\frac{e^2\sqrt{Q+1}}{R^3}}$
Problem 6.
Express the following in log form, with no fractions or powers inside the log.
(a) $V=\frac{\pi{}h}{4}\left(D-h\right)\left(D+h\right)$
(b) $P=\frac{1}{16}(2d-1)^2N\sqrt{S}$
Problem 7.
Solve the following equations using logarithm theorems.
(a) $\log{(5x-1)}=2$
(b) $\log_2{(x+1)}-\log_2{(x-4)}=3$
(c) $\log_6{(x+4)}+\log_6{(x-2)}=\log_6{(4x)}$ where $x>0$
$\Rightarrow{}\ \ x^2-2x-8=0$
$\Rightarrow{}\ \ \ (x-4)(x+2)=0\ \ \ x_1=4 \text{ and } x_2=-2\ \ \ \Rightarrow{}\ \ \ \boxed{x=4}$
Problem 8.
Determine the domain and range of the functions defined below.
(a) $f(x)=\cos^2{(x)}$
$\Rightarrow{} \boxed{ \text{Range of f }\left(x\right)\Rightarrow{}[0,1].}$
(b) $f(x)={-x}^2+5x-6$
$\Rightarrow{}\boxed{\text{Range of } f(x) \Rightarrow{} \text{, function has one maximum.} \Rightarrow{}\quad y\leq \frac{1}{4}}$
(c) $f(x)=-x^3+4x^2-4x\ $
$\Rightarrow{}\boxed{\text{Range of f}\left(x\right)\Rightarrow{}\ \ (-\infty{},\infty{})}$
(d) $f\left(x\right)=$exp$(1/x)$
$y=e^{\frac{1}{x}}\ \Rightarrow{}$ 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) $f\left(x\right)=\frac{1}{e^x+1}$
$y=\frac{1}{e^x+1}\ \Rightarrow{}$ 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) $\log_2{15}$
(b) $\log_{20}{17}$
(c) $\log_3{16}$
Problem 10.
Determine if the function $f(x)$ is odd, even, or neither.
(a) $f\left(x\right)=\ x^2\sin{(2x)}$
(b) $f\left(x\right)=3\sin{\left(x\right)}\cos{\left(4x\right)}$
(c) $f\left(x\right)=x^3e^3$
(d)
(e)
(f)
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
(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.
$\Rightarrow 4(2)^2 + b(2) - 6 = 0 \Rightarrow \boxed{b = - 5}$
$\Rightarrow$ 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.
$\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 $y \geqslant 0$.
(e) Find the domain (using the set notation)
(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.
(b) State, with justification, how the curve approaches the horizontal asymptote for large positive and large negative values of x.
$\text{Large negative} \ x, \boxed{ \ y \rightarrow{3}\text{ from above}}$
(c) Sketch the curve
(d) Solve the inequality $\frac{3x^2 - 9}{x^2 + 3x -4} \geqslant 0. $
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.
$a + 1 = 4 \Rightarrow a = 3$
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.
$b(-1)+2 = 0 \Rightarrow b = 2$
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 $P$.
$\frac{\partial y}{\partial x} = 3x^2-6x-9$
$\frac{\partial ^{2}y}{\partial x^{2}} = 6x-6$
$\Rightarrow\quad\ x = 1, \therefore y = -8$ 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 $-f(x) = f(-x)$
$f(-x) = (-x)^3-12(-x) = -x^3+12x = -f(x)$
$\boxed{\text{Therefore it is odd around the point P.}}$
Answers
For Printing
Revision Questions
The questions included are optional, but here if you want some extra practice.
- Engineering Mathematics 7th edition, Stroud and Dexter : Pages 63-100, 315-343, 353-381, 688-707, 727-728
- 3D Curve Sketching : More of an extension, but with WolframAlpha, Sam could throw anything…