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Week 2 (lectures 3 and 4): In this question, you will look at functions, surjectivity, some graph-plotting techniques, function composition and differentiating via the chain-rule, and Taylor polynomials.

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As with all questions, there may be parts where you can choose to \"Show steps\". This may give a hint, or it may present sub-parts which will help you to solve that part of the question. Furthermore, remember to always press the \"Show feedback\" button at the end of each part. Sometimes, helpful feedback will be given here, and often it will depend on how correctly you have answered and will link to other parts of the question. Hence, always retry the parts until you obtain full marks, and then look at the feedback again.

Keep in mind that in order to see the feedback for a particular part of a question, you must provide a full (but not necessarily correct) answer to that part. Do not worry though, as you can look at the feedback and then ammend your answer accordingly.

Furthermore, as with all questions, choosing to reveal the answers will only show you the answers which change every time the question is loaded (i.e. answers to randomised questions); the fixed answers will not be revealed.

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What is the definition of a function?
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Let $X$ and $Y$ be sets, and let $f: X \\longrightarrow Y$ be a function. What does it mean for $f$ to be surjective?
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Given two sets $X$ and $Y$, a function $f: X \\longrightarrow Y$ is a rule that assigns, to each $y \\in Y$, a unique element $x \\in X$.

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Given two sets $X$ and $Y$, a function $f: X \\longrightarrow Y$ is a rule that assigns, to each $y \\in Y$, at least one element $x \\in X$.

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Given two sets $X$ and $Y$, a function $f: X \\longrightarrow Y$ is a rule that assigns, to each $x \\in X$, a unique element $f(x) \\in Y$.

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Given two sets $X$ and $Y$, a function $f: X \\longrightarrow Y$ is a rule that assigns, to each $x \\in X$, at least one element $f(x) \\in Y$.

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Given two sets $X$ and $Y$, a function $f: X \\longrightarrow Y$ is a rule that assigns, to each $x \\in X$, a unique element $f(x) \\in Y$. Furthermore, for each $y \\in Y$, there must be at least one $x \\in X$ with $y = f(x)$.

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For all $x_1 , x_2 \\in X$, if $x_1 \\neq x_2$ then $f(x_1 ) \\neq f(x_2 )$.

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For all $y \\in Y$ there is a unique $x \\in X$ such that $y = f(x)$.

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For all $x \\in X$, there is a unique $y \\in Y$ such that $y = f(x)$.

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For all $x \\in X$, there is at least one $y \\in Y$ such that $y = f(x)$.

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For all $y \\in Y$ there is at least one $x \\in X$ such that $y = f(x)$.

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Let us try to write $\\displaystyle{x^2 - \\simplify{2*{Centre_x}}x +y^2 -\\simplify{2*{Centre_y}}y + \\simplify{{Centre_x}^2 +{Centre_y}^2 - {Radius}^2}=0}$ in the form $(x-a)^2 + (y-b)^2 = r^2$. We can determine the centre and radius easily. Indeed, the centre is $(a,b)$ and the radius is $r$.

First, expand $(x-a)^2 + (y-b)^2 = r^2$.

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So, by comparing the coefficients of the $x$ term, we see that $a=$

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So, by comparing the coefficients of the $y$ term, we see that $b=$

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Finally, by comparing the constant terms (including $a^2$ and $b^2$), we see that $r=$

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Determine the centre and radius of the circle with the following equation:
$\\simplify[all, !noLeadingMinus]{x^2 - {2*{Centre_x}}x +y^2 -{2*{Centre_y}}y + {{Centre_x}^2 +{Centre_y}^2 - {Radius}^2}=0}$

Centre: $($[] $,$ [] $)$

It may be helpful for you to draw the circle, which should be easier now that you know its radius and centre, from part b.

For the first part: Recall that for a function $g$, each $x$ in the domain has only one corresponding $g(x)$ in the codomain. That means that any vertical line will pass through at most one point on the graph of $g$. We must find a value of $d$ such that any vertical line passes through $S_G \\cup L_d$ at most once.

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For the second part: Now that we have our function, we want to restrict its codomain so that it is a surjective function. Recall that for a general function, not all elements in the codomain are necessarily mapped to. With a surjective function though, it must be the case that all elements in the codomain are mapped to.

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For example, consider the function $f: \\mathbb{R} \\longrightarrow \\mathbb{R}$ defined by $f(x) = 1$ for all $x \\in \\mathbb{R}$. Clearly there are elements in the codomain, $\\mathbb{R}$, which are not mapped to, such as $2,3,$ or $4$. Hence this function is not surjective. However, we can restrict the codomain to make it surjective. That is, following function is surjective: $f: \\mathbb{R} \\longrightarrow \\{ 1 \\}$ defined by $f(x) = 1$ for all $x \\in \\mathbb{R}$. We simply restricted the codomain from $\\mathbb{R}$ to $\\{ 1 \\}$.

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Returning to the question, it is your task to restrict the codomain so that function is surjective.

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Consider the subset $S_G \\subseteq \\mathbb{R}$ consisting of all points $(x,y) \\in \\mathbb{R}$ which satisfy the cartesian equation $\\simplify[all, !noLeadingMinus]{x^2 - {2*{Centre_x}}x +y^2 -{2*{Centre_y}}y + {{Centre_x}^2 +{Centre_y}^2 - {Radius}^2}=0}$.

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Now let $d \\in \\mathbb{R}$, and define the subset $L_d \\subseteq \\mathbb{R}$ to be the \"half-plane\" $\\{ (x,y) \\in \\mathbb{R} : y \\leq d \\}$.

What value must we take for $d$ so that (1) $s_g \\cap L_d$ corresponds to the graph of a function (we will call this function $f$), and (2) this graph is as large as possible? $d =$ [].

We can now express this function as $f: [ \\simplify{{DomainLimitLeft}} , \\simplify{{DomainLimitRight}}] \\longrightarrow \\Big(- \\infty,$[]$\\Big]$.

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We now wish to find a number $c \\in \\mathbb{R}$ such that restricting the codomain to $\\Big[c,$[]$\\Big]$ will make the function $f$ surjective.

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What value do we take for $c$? $c=$ [].

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Hint for part iii: We have two functions which are formed by composition, namely $f \\circ g$ and $g \\circ f$, and you are asked to find their derivative. This implies that we should use the chain rule, which we have from part i. The following is an example of an application of the chain rule:

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Let $h(x)= e^x$ and let $k(x) = x^2 +x$.
Then $(h \\circ k ) (x) = e^{x^2 + x}$.

The chain rule tells us that $\\frac{\\mathrm{d} \\big( (h \\circ k) (x) \\big)}{\\mathrm{d} x} = \\frac{\\mathrm{d}(h(k))}{\\mathrm{d} k} \\cdot \\frac{\\mathrm{d}(k(x))}{\\mathrm{d} x}$.
Now, $h(k) = e^k$ and so $\\frac{\\mathrm{d}\\big( (h \\circ k) (x) \\big)}{\\mathrm{d} k} = e^k$.
Also, $\\frac{\\mathrm{d}(k(x))}{\\mathrm{d} x} = 2x+1$.
Hence, $\\frac{\\mathrm{d} \\big( (h \\circ k) (x) \\big)}{\\mathrm{d} x} = (2x+1) e^k$.

Finally, we can express $k$ in terms of $x$ by using $k(x) = x^2 + x$, to get $\\frac{\\mathrm{d} \\big( (h \\circ k) (x) \\big)}{\\mathrm{d} x} = (2x+1) e^{x^2 +x}$.

Now try something similar for part iii.

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This part of the question involves differentiation via the chain rule.

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i) Suppose we have functions $f: \\mathbb{R} \\longrightarrow \\mathbb{R}$ and $g: \\mathbb{R} \\longrightarrow \\mathbb{R}$. What does the chain rule say?
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ii) Now, let $f(x)= x^2 + \\simplify{{a}}$ and let $g(x)= \\simplify{{b}} \\cos (x) + \\simplify{{c}}$. Express, in terms of $x$, the compositions $f \\circ g$ and $g \\circ f$.
(When entering an expression, please use * for multiplication. For example write x * sin(x), and not x sin(x); the latter may not always be interpreted correctly by the system. Also, if you wish to write $\\cos^2 (x)$, then please write cos(x)^2, and not cos^2(x). This applies similarly to $\\sin ,\\tan$ etc.. )
$(f \\circ g) (x) =$ []
$(g \\circ f) (x) =$ []

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iii) Using the chain rule, calculate $\\frac{\\mathrm{d} \\big( (f \\circ g) (x) \\big)}{\\mathrm{d} x}$ and $\\frac{\\mathrm{d} \\big( (g \\circ f) (x) \\big)}{\\mathrm{d} x}$.
$\\frac{\\mathrm{d} \\big( (f \\circ g) (x) \\big)}{\\mathrm{d} x} =$ []
$\\frac{\\mathrm{d} \\big( (g \\circ f) (x) \\big)}{\\mathrm{d} x} =$ []

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$\\frac{\\mathrm{d}\\big(f(x) + g(x) \\big)}{\\mathrm{d} x} = \\frac{\\mathrm{d}(f(x))}{\\mathrm{d} x} + \\frac{\\mathrm{d}(g(x))}{\\mathrm{d} x}$

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$\\frac{\\mathrm{d} \\big( (f \\circ g) (x) \\big)}{\\mathrm{d} x} = \\frac{\\mathrm{d}(f(g))}{\\mathrm{d} g} \\cdot \\frac{\\mathrm{d}(g(x))}{\\mathrm{d} x}$

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$\\frac{\\mathrm{d} (f(x) \\cdot g(x))}{\\mathrm{d} x} = f(x) \\cdot \\frac{\\mathrm{d}(g(x))}{\\mathrm{d} x} + \\frac{\\mathrm{d}(f(x))}{\\mathrm{d} x} \\cdot g(x)$

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$\\frac{\\mathrm{d} \\big( f(x) /g(x) \\big)}{\\mathrm{d} x} = \\frac{\\frac{\\mathrm{d} (f(x))}{\\mathrm{d} x} \\cdot g(x) - f(x) \\cdot \\frac{\\mathrm{d} (g(x))}{\\mathrm{d} x}}{g(x)^2 }$

Hint for part i: This was explained in chapter 4 of the lecture notes.

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Hint for part part ii: The (correct) answer from part i tells you the general form of the taylor polynomial. To apply it to this part of the question, you will need to calculate the derivative of $h(x) = (\\simplify{{b}} \\cos (x) + \\simplify{{c}})^2 + \\simplify{{a}}$. However, you already did this in part d.

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This part of the question involves Taylor Polynomials.

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i) Let $h: \\mathbb{R} \\longrightarrow \\mathbb{R}$ be a function which is differentiable at least twice, and let $a \\in \\mathbb{R}$. Then, the Taylor polynomial of $h$ about $a$ of degree $1$ is
[]

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ii) Now consider the function $h(x) = (\\simplify{{b}} \\cos (x) + \\simplify{{c}})^2 + \\simplify{{a}}$. Notice that this is the same as $(f \\circ g)(x)$ from part d.

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What is the Taylor polynomial of $h$ about $0$ of degree $1$? (If you wish to write $\\pi$, then simply write \"pi\", without the quotation marks.)
$P_2 (x) =$ [] .

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What is the Taylor polynomial of $h$ about $\\frac{\\pi}{2}$ of degree $1$?
$P_2 (x) =$ [] .

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$P_1 (x) = h'(0) x+h(0)$

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$P_1 (x) =h'(a) x +h(a)$

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$P_1 (x) =h'(a) (x-a) +h(a)$

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$P_1 (x) =h'(a) (x-a) +h(0)$

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$P_2 (x) =h'(0) (x-a) +h(0)$

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This is the question for week 2 of the MA100 course at the LSE. It looks at material from chapters 3 and 4.

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This is the left limit of the domain of the function that corresponds to the lower semicircle.

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This is the right limit of the domain of the function that corresponds to the lower semicircle.

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Centre_x is the x coordinate of the centre of the circle.

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Centre_y is the y coordinate of the centre of the circle.

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\"Radius\" is the radius of the circle. We want the the circle to have two intersections with each axis. Hence the definition of its value.

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\"d\" is a constant in a function.

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\"c\" is a constant in a function.

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\"b\" is a constant in a function.

", "group": "Ungrouped variables"}, "a": {"definition": "random(2..10)", "name": "a", "templateType": "anything", "description": "

\"a\" is a constant in a function.

", "group": "Ungrouped variables"}}, "variable_groups": [], "advice": "", "extensions": [], "type": "question", "contributors": [{"name": "Michael Yiasemides", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1440/"}]}]}], "contributors": [{"name": "Michael Yiasemides", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1440/"}]}