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Week 4 (lectures 7 and 8): In this question, you will look at graph sketching, local inverses, and integration.

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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.

", "variables": {"PolySign": {"description": "

This is a scalar that we will multiply our polynomial by. If PolySign is -1, then the graph of our polynomial will be below the x-axis. If PolySign is 1, then the graph will be above the x-axis. Ultimately, this gives some randomness to the classification of the stationary points of our polynomial. See the question description for the full explanation of our polynomial.

", "group": "Variables for parts a to c", "name": "PolySign", "definition": "random(-1,1)", "templateType": "anything"}, "LocInvRightDom": {"description": "

This is the right limit of the of the domain of the local inverse. It depends on the value PartitionNumber, since it is PartitionNumber (via LocInvPoint) that tells the student where we want the local inverse to be taken.

", "group": "Variables for parts a to c", "name": "LocInvRightDom", "definition": "switch(PartitionNumber = 1, -a , PartitionNumber = 2, 0 , PartitionNumber = 3 , a , PartitionNumber = 4, 2*a, '')", "templateType": "anything"}, "YSign": {"description": "

This is the sign that goes in front of y in the local inverse.

", "group": "Variables for parts a to c", "name": "YSign", "definition": "if(PolySign = -1 , \"-\" , \"\")", "templateType": "anything"}, "LocInvPoint": {"description": "

This is the point at which we will ask the student to find a local inverse. It is dependent on the PartitionNumber.

", "group": "Variables for parts a to c", "name": "LocInvPoint", "definition": "switch(PartitionNumber = 1, -a-1 , PartitionNumber = 2, -a/2 , PartitionNumber = 3 , a/2 , PartitionNumber = 4, a+1, '')", "templateType": "anything"}, "c": {"description": "

A constant that will appear in our function for part d.

", "group": "Variables for part d and onwards", "name": "c", "definition": "random(1..5)", "templateType": "anything"}, "d": {"description": "

A constant that will appear in our function for part d.

", "group": "Variables for part d and onwards", "name": "d", "definition": "random(1..5)", "templateType": "anything"}, "a": {"description": "

The stationary points of our polynomial will be -a,0,a. See question description for the full explanation of our polynomial.

", "group": "Variables for parts a to c", "name": "a", "definition": "random(2..6)", "templateType": "anything"}, "LocInvCodom": {"description": "

One of the limits of the codomain of the local inverse will be PolySign*b. LocInvCodom is the other limit. It depends on the value PartitionNumber, since it is PartitionNumber (via LocInvPoint) that tells the student where we want the local inverse to be taken.

", "group": "Variables for parts a to c", "name": "LocInvCodom", "definition": "switch(PartitionNumber = 1, PolySign*(9*a^4 + b) , PartitionNumber = 2, PolySign*(a^4 + b) , PartitionNumber = 3 , PolySign*(a^4 + b) , PartitionNumber = 4, PolySign*(9*a^4 + b) , '')", "templateType": "anything"}, "LocInvLeftDom": {"description": "

This is the left limit of the of the domain of the local inverse. It depends on the value PartitionNumber, since it is PartitionNumber (via LocInvPoint) that tells the student where we want the local inverse to be taken.

", "group": "Variables for parts a to c", "name": "LocInvLeftDom", "definition": "switch(PartitionNumber = 1, -2*a , PartitionNumber = 2, -a , PartitionNumber = 3 , 0 , PartitionNumber = 4, a, '')", "templateType": "anything"}, "RandomStatPoint": {"description": "

A random stationary point of our function.

", "group": "Variables for parts a to c", "name": "RandomStatPoint", "definition": "random(a , -1*a , 0)", "templateType": "anything"}, "FirstSign": {"description": "

When attempting to find the inverse of our function, we obtain two \"plus-or-minuses\" in the expression. The correct combination depends on the partition number. For the first \"plus-or-minus\" we have a plus when PartitionNumber = 3 or 4, otherwise we have a minus.

", "group": "Variables for parts a to c", "name": "FirstSign", "definition": "switch(PartitionNumber = 3 , \"+\" , PartitionNumber = 4 , \"+\" , \"-\")", "templateType": "anything"}, "k": {"description": "

A constant that will appear in our function for part e.

", "group": "Variables for part d and onwards", "name": "k", "definition": "random(2..6)", "templateType": "anything"}, "OppFirstSign": {"description": "

This is the opposite sign to FirstSign. We need it in one of the questions where we ask students to choose the local inverse from a set of answers.

", "group": "Variables for parts a to c", "name": "OppFirstSign", "definition": "switch(PartitionNumber = 3 , \"-\" , PartitionNumber = 4 , \"-\" , \"+\")", "templateType": "anything"}, "g": {"description": "

A constant that will appear in our function for part d.

", "group": "Variables for part d and onwards", "name": "g", "definition": "random(2..6)", "templateType": "anything"}, "PartitionNumber": {"description": "

the domain of our function can be partitioned (more or less) into 4 parts, where the function is invertible when restricted to one of these parts, and each part is as large a possible. PartitionNumber is the parts for which we will require the student to find a local inverse.

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The parts are (-\\infty , -a], [-a , 0], [0 , a], [a , \\infty), corresponding to PartitionNumber = 1,2,3,4 respectively.

", "group": "Variables for parts a to c", "name": "PartitionNumber", "definition": "random(1..4)", "templateType": "anything"}, "b": {"description": "

This is a constant that is added to our polynomial to ensure that the polynomial does not cross the x-axis. See the question description for the full explanation of our polynomial.

", "group": "Variables for parts a to c", "name": "b", "definition": "random(1..5)", "templateType": "anything"}, "OppSecondSign": {"description": "

This is the opposite sign to SecondSign. We need it in one of the questions where we ask students to choose the local inverse from a set of answers.

", "group": "Variables for parts a to c", "name": "OppSecondSign", "definition": "switch(PartitionNumber = 1 , \"-\" , PartitionNumber = 4 , \"-\" , \"+\")", "templateType": "anything"}, "f": {"description": "

A constant that will appear in our function for part d.

", "group": "Variables for part d and onwards", "name": "f", "definition": "random(1..5)", "templateType": "anything"}, "SecondSign": {"description": "

When attempting to find the inverse of our function, we obtain two \"plus-or-minuses\" in the expression. The correct combination depends on the partition number. For the second \"plus-or-minus\" we have a plus when PartitionNumber = 1 or 4, otherwise we have a minus.

", "group": "Variables for parts a to c", "name": "SecondSign", "definition": "switch(PartitionNumber = 1 , \"+\" , PartitionNumber = 4 , \"+\" , \"-\")", "templateType": "anything"}, "h": {"description": "

A constant that will appear in our function for part e.

", "group": "Variables for part d and onwards", "name": "h", "definition": "random(2..6)", "templateType": "anything"}, "j": {"description": "

A constant that will appear in our function for part e.

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Comment for part ii: Let $a,b,c \\in \\mathbb{R}$ and suppose we have a polynomial $f: \\mathbb{R} \\longrightarrow \\mathbb{R}$ defined by $f(z) = az^2 + bz + c$ for all $z \\in \\mathbb{R}$. The quadratic formula tells us what the roots of the polynomial are: $\\frac{-b \\pm \\sqrt{b^2 -4ac}}{2a}$. Notice that we have real-valued roots if and only if $b^2 - 4ac \\geq 0$ (otheriwse $\\sqrt{b^2 -4ac}$ would be a complex number). That is, our polynomial intercepts the $x$-axis if and only if $b^2 -4ac \\geq 0$. This is why we call $b^2 - 4ac$ the discriminant: It allows us to detrmine if our polynomial intercepts the $x$-axis.

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Hint for part iii: Recall that we may be able to classify a stationary point by evaluating the second derivative of the function at that point. For example, if $f''(s_1) > 0$ then $s_1$ is a minimum point; and if $f''(s_1) < 0$ then $s_1$ is a maximum point.

"}], "variableReplacements": [], "prompt": "

Consider the function $f:[\\simplify{-2*{a}} , \\simplify{2*{a}}] \\longrightarrow \\mathbb{R}$ defined by $f(x)= \\simplify[all , !noLeadingMinus]{{PolySign}* x^4 -2*{PolySign}* {a}^2 * x^2 + {PolySign}*{a}^4 + {PolySign}*{b}}$ for all $x \\in [\\simplify{-2*{a}} ,\\simplify{2*{a}}]$.

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First, we would like to find the axes intercepts of this function, and determine and classify the stationary points. This will make it easier for you to draw the graph of the function, which will help in the later parts of this question.

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i) What is the y-intercept of the function? [[0]]

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ii) Note that we can view our function as a polynomial in $x^2$. That is, a polynomial of the form $a (x^2)^2 + b (x^2) + c$ for some $a,b,c \\in \\mathbb{R}$. What is the discriminant of this polynomial? [[1]] (Press the \"Show steps\" button below for an explanation of the discriminant.)

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Hence we can see that this polynomial (i.e. our function) intercepts the $x$-axis [[2]] times.

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iii) Now we will find and classify the stationary points. First find the the first deivative and second derivative of $f$.
(If you wish to write something like $2x^2$ then please write 2*x^2 and not 2x^2, as the latter may not be correctly interpreted by the system.)
$f'(x)=$ [[3]]
$f''(x)=$ [[4]]

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We can now use $f'(x)$ to find the stationary points. For this function there are three stationary points which we will call $s_1$, $s_2$, and $s_3$, with $s_1 < s_2 < s_3$.
$s_1=$ [[5]]
$s_2=$ [[6]]
$s_3=$ [[7]]

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Now we will evaluate the second derivative at these stationary points, to determine whether they are maximum, minimum, or inflection points.

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$f''(s_1) =$ [[8]] and so $s_1$ is a/an [[9]] point.
$f''(s_2) =$ [[10]] and so $s_2$ is a/an [[11]] point.
$f''(s_3) =$ [[12]] and so $s_3$ is a/an [[13]] point.

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1

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2

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{if(PolySign > 0 , \"minimum\" , \"maximum\")}

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{if(PolySign > 0 , \"maximum\" , \"minimum\")}

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inflection

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{if(PolySign > 0 , \"maximum\" , \"minimum\")}

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{if(PolySign > 0 , \"minimum\" , \"maximum\")}

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inflection

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{if(PolySign > 0 , \"minimum\" , \"maximum\")}

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{if(PolySign > 0 , \"maximum\" , \"minimum\")}

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inflection

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Notice that our function is not bijective: We can easily see from our graph that it is not injective as there are horizontal lines which intersect the graph at more than one point.

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Hence, we can say that our function has no inverse. However there may be local inverses at given points. (We recommend that you look at section 7.4 of the lecture notes for an explanation of local inverse, before continuing.)

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[[0]]

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Does a local inverse for $f$ exist at the point $x_0 = \\var{{RandomStatPoint}}$.

", "

Does a local inverse for $f$ exist at the point $x_1 =$ {random(-a-1 ,-a/2 , a/2 , a+1)}

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Let $x_2 = \\var{{LocInvPoint}}$. We want to find the local inverse of $f$ at this point. First we need to find $A \\subseteq [\\simplify{-2*{a}} , \\simplify{2*{a}}]$ and $B \\subseteq \\mathbb{R}$ such that $f: A \\longrightarrow B$ is invertible, $x_2$ is an interior point of $A$, and $f(x_2)$ is an interior point of $B$. Furthermore, we would like $A$ and $B$ to be as large as possible. Fill in the following gaps to demonstrate what $A$ and $B$ are.

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$A = \\big[$[[0]] , [[1]]$\\big]$.
$B = \\big[$[[2]] , [[3]]$\\big]$.

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Now, if we let $y = f(x) = \\simplify[all , !noLeadingMinus]{{PolySign}* x^4 -2*{PolySign}* {a}^2 * x^2 +{PolySign}*{a}^4 +{PolySign}*{b}}$ for all $x \\in[\\simplify{-2*{a}} ,\\simplify{2*{a}}]$, and try to obtain $x$ in terms of $y$, we would get $x = \\pm \\sqrt{\\simplify{{a}^2} \\pm \\sqrt{\\var{{YSign}} y-\\var{{b}}}}$. There are two occurences of $\\pm$, meaning there are four possible combinations in total. Only one of these combinations will represent the local inverse of $f$ at $x_2 = \\var{{LocInvPoint}}$. The question is: Which one?

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There is more than one way to go about answering this. Here is one suggestion: Choose a point $y$ in the interior of $B$, and evaluate $x = \\pm \\sqrt{\\simplify{{a}^2} \\pm \\sqrt{y-\\var{{b}}}}$ for each possible combination. Only one combination  will give you a value which lies in $A$. It is therefore this combination that tells us what the local inverse of $f$ at $x_2 = \\var{{LocInvPoint}}$ is.

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The local Inverse of $f$ at $x_2 = \\var{{LocInvPoint}}$ is defined by $f^{-1} : B \\longrightarrow A$, where for each $y \\in B$ we have
[[4]]

$y = \\simplify{{FirstSign}} \\sqrt{\\simplify{{a}^2} \\simplify{{SecondSign}} \\sqrt{\\var{{YSign}}y-\\var{{b}}}}$

", "

$y = \\simplify{{FirstSign}} \\sqrt{\\simplify{{a}^2}\\simplify{{OppSecondSign}}\\sqrt{\\var{{YSign}}y-\\var{{b}}}}$

", "

$y = \\simplify{{OppFirstSign}} \\sqrt{\\simplify{{a}^2}\\simplify{{SecondSign}}\\sqrt{\\var{{YSign}}y-\\var{{b}}}}$

", "

$y = \\simplify{{OppFirstSign}} \\sqrt{\\simplify{{a}^2}\\simplify{{OppSecondSign}}\\sqrt{\\var{{YSign}}y-\\var{{b}}}}$

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The remainder of this question involves integration.

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Consider te function $g:\\mathbb{R} \\longrightarrow \\mathbb{R}$ defined by $g(x) = \\var{{c}} e^{\\var{{d}}x}$ for $x \\in (-\\infty , 0)$, and $g(x) = \\frac{\\var{{f}}}{(x+1)^2} - \\frac{\\var{{g}}}{(x+1)^3}$ for $x \\in [0, \\infty)$.

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Complete the following gaps to calculate $\\int_{-\\infty}^{\\infty} g(x) \\mathrm{d} x$. (For the first two gaps you will need to calculate the primitive functions (these are described at the start of chapter 8 of the notes), and for the third and fourth gaps you will need to evaluate these functions at the limits.)

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$\\int_{-\\infty}^{\\infty} g(x) \\mathrm{d} x = \\int_{-\\infty}^{0} \\var{{c}} e^{\\var{{d}}x} \\mathrm{d} x + \\int_{0}^{\\infty} \\frac{\\var{{f}}}{(x+1)^2} - \\frac{\\var{{g}}}{(x+1)^3} \\mathrm{d} x = \\Big[$ [[0]] $\\Big]_{-\\infty}^{0} + \\Big[$ [[1]] $\\Big]_{0}^{\\infty} =$ [[2]] $+$ [[3]] $=$ [[4]].

Comment: You may be concerned that we use $h(u)$ instead of $h(x)$, but this is nothing to worry about. $x$ and $u$ are just \"dummy-variables\". Ultimately all that matters is the rule that defines the function $h$. For example, if we have a function $k:\\mathbb{R} \\longrightarrow \\mathbb{R}$ then the statements \"$k(x) = x+1$ for all $x \\in \\mathbb{R}$\", \"$k(u) = u+1$ for all $u \\in \\mathbb{R}$\", and \"$k(y) = y+1$ for all $y \\in \\mathbb{R}$\" all say the same thing and are equivalent.

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Hint: If $x \\in [0, \\infty)$ then the integral $\\int_{-\\infty}^{x} h(u) \\mathrm{d} u$ ranges over $(-\\infty , 0)$ and $[0,x] \\subseteq [0, \\infty)$, and $h(u)$ is defined differently on these two sets. Hence, we must split the the integral into two parts, $\\int_{-\\infty}^{0} h(u) \\mathrm{d} u + \\int_{0}^{x} h(u) \\mathrm{d} u$, and apply the corresponding definition of $h(u)$ to each one.

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If $x \\in (-\\infty , 0)$, then the integral $\\int_{-\\infty}^{x} h(u) \\mathrm{d} u$ ranges over $(-\\infty , x) \\subseteq (-\\infty , 0)$ only, and there is only one definition of $h(u)$ on this set. Hence, there is no need to split the integral in this case.

"}], "variableReplacements": [], "prompt": "

Now consider the function $h:\\mathbb{R} \\longrightarrow \\mathbb{R}$ defined by $h(x) = \\frac{1}{(x-\\var{{h}})^2}$ for $x \\in (-\\infty , 0)$, and $h(x) = \\var{{j}} \\cos ( \\var{{k}} x )$ for $x \\in [0 , \\infty)$.

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Also, define the function $H: \\mathbb{R} \\longrightarrow \\mathbb{R}$ by $H(x) = \\int_{-\\infty}^{x} h(u) \\mathrm{d} u$ for all $x \\in \\mathbb{R}$.

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Fill in the gaps below to find an explicit expression for $H(x)$ when $x \\in (-\\infty , 0)$, and another explicit expression when $x \\in [0 , \\infty)$. For the first, third, and fourth gaps you will need to enter the appropriate expression for $h(u)$; press the \"Show steps\" button below for a hint if you need one. (If you wish to write something like $6 \\cos(u)$ then please write 6*cos(u) and not 6cos(u), as the system may not correctly interpret the latter.)

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When $x \\in (-\\infty , 0)$ we have that
$H(x) = \\int_{-\\infty}^{x} h(u) \\mathrm{d} u = \\int_{-\\infty}^{x}$ [[0]] $\\mathrm{d} u =$ [[1]] .

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When $x \\in [0 , \\infty)$ we have that
$H(x) = \\int_{-\\infty}^{x} h(u) \\mathrm{d} u = \\int_{-\\infty}^{0}$ [[2]] $\\mathrm{d} u + \\int_{0}^{x}$ [[3]] $\\mathrm{d} u =$ [[4]] + [[5]] .

For parts a to c, we used a polynomial defined as m*(x^4 - 2a^2  x^2 + a^4 + b), where the variables \"a\" and \"b\" are randomly chosen from a set of reaosnable size, and the variable $m$ is randomly chosen from the set {+1, -1}. We can easily see that this polynomial has stationary points at -a, 0, and a. We introduced the variable \"m\" so that these stationary points would not always have the same classification. The variable \"b\" is always positive, and so this ensures that our polynomial does not cross the x-axis. The first and second derivatives; stationary points; the evaluation of the second derivative at the stationary points; the classification of the stationary points; and the axes intercepts can all be easily expressed in terms of the variables \"a\", \"b\", and \"m\". Indeed, this is what we did to mark the student's answers.