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Sketch a function graph by using fist and second derivatives.

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Consider the function \$ f: \\mathbb{R} \\to \\mathbb{R} : x \\mapsto \\ln{ \\frac{\\var{c} \\cdot x-\\var{a}}{x-\\var{b}} }\\, . \$

", "advice": "

a) In order to obtain the domain of \$ f \$ , we need to solve the inequality \\[ \\frac{\\var{c}\\cdot x-\\var{a}}{x-\\var{b}} > 0 \\]

Since the sign of \$ \\frac{\\var{c}\\cdot x-\\var{a}}{x-\\var{b}} \$ is given by

\\[
\\begin{array}{ c | c c c c c}
x   &            & \\simplify[all,fractionNumbers]{{nulp1}} & & \\simplify[all,fractionNumbers]{{nulp2}} &    \\\\
    \\hline
\\frac{\\var{c}\\cdot x-\\var{a}}{x-\\var{b}} & + & \\var{teken1} & - & \\var{teken2} & +    \\\\
\\end{array}
\\]
we get
\\[ dom f = \\left ] - \\infty , \\simplify[all,fractionNumbers]{{nulp1}} \\right[ \\quad \\cup \\quad \\left] \\simplify[all,fractionNumbers]{{nulp2}} , +\\infty \\right[ .\\]

b) Since \$ \\mathrm{D} \\ln x = \\frac{1}{x} \$ , the chain rule leads to
\\[ \\mathrm{D} \\ln{\\left( \\frac{\\var{c}\\cdot x-\\var{a}}{x-\\var{b}}\\right) } = \\frac {x-\\var{b}} {\\var{c}\\cdot x-\\var{a}} \\cdot \\mathrm{D} \\frac{\\var{c}\\cdot x-\\var{a}}{x-\\var{b}} , \\]
i.e.
\\begin{align} f'(x) &= \\frac {x-\\var{b}} {\\var{c}\\cdot x-\\var{a}} \\cdot \\frac{(x-\\var{b})\\cdot \\var{c} - (\\var{c}\\cdot x-\\var{a}) \\cdot 1}{(x-\\var{b})^2} \\\\
&= \\frac{\\var{c}\\cdot (x-\\var{b}) - (\\var{c}\\cdot x-\\var{a}) }{(\\var{c}\\cdot x-\\var{a}) \\cdot (x-\\var{b})} \\\\
&= \\frac{\\var{a-b*c} }{(\\var{c}\\cdot x-\\var{a}) \\cdot (x-\\var{b})}
\\end{align}

Remark: you can get the same result (making abstraction of the domain) by using
\\[ \\ln{\\left( \\frac{\\var{c}\\cdot x-\\var{a}}{x-\\var{b}}\\right) } = \\ln{\\left( {\\var{c}\\cdot x-\\var{a}}\\right) } - \\ln{\\left( {x-\\var{b}}\\right) } , \\]
such that
\\begin{align}
\\mathrm{D} \\ln{\\left( \\frac{\\var{c}\\cdot x-\\var{a}}{x-\\var{b}}\\right) } &= \\mathrm{D} \\left( \\ln { \\left( {\\var{c}\\cdot x-\\var{a}} \\right)} \\right) - \\mathrm{D} \\left( \\ln { \\left( {x-\\var{b}} \\right)} \\right) \\\\
&= \\frac{1}{\\var{c}\\cdot x-\\var{a}} \\cdot \\var{c}- \\frac{1}{x-\\var{b}}\\\\
&= \\frac{ \\var{c} \\cdot ({x-\\var{b}}) - (\\var{c} \\cdot x-\\var{a}) }{(\\var{c} \\cdot x-\\var{a}) \\cdot (x-\\var{b})} \\\\
&= \\frac{\\var{a-b*c} }{(\\var{c} \\cdot x-\\var{a}) \\cdot (x-\\var{b})}
\\end{align}

c) The second order derivative can be calculated as follows:
\\[ \\begin{align}
\\mathrm{D} \\frac{\\var{a-b*c} }{(\\var{c} \\cdot x-\\var{a}) \\cdot (x-\\var{b})} &= {\\var{a-b*c}} \\cdot \\mathrm{D} {\\left( (\\var{c} \\cdot x-\\var{a}) \\cdot (x-\\var{b}) \\right) ^{-1}}\\\\
&= {\\var{-a+b*c}} \\cdot {\\left( (\\var{c} \\cdot x-\\var{a}) \\cdot (x-\\var{b}) \\right) ^{-2}} \\cdot \\mathrm{D} {\\left( (\\var{c} \\cdot x-\\var{a}) \\cdot (x-\\var{b}) \\right) }\\\\
&= {\\var{-a+b*c}} \\cdot {\\frac {1}{\\left( (\\var{c} \\cdot x-\\var{a}) \\cdot (x+\\var{b}) \\right) ^{2}}} \\cdot {\\left( \\var{c} \\cdot (x-\\var{b}) + (\\var{c} \\cdot x-\\var{a}) \\right) }\\\\
&= {\\var{-a+b*c}} \\cdot {\\frac {1}{\\left( (\\var{c} \\cdot x-\\var{a}) \\cdot (x+\\var{b}) \\right) ^{2}}} \\cdot {\\left( \\var{2*c} \\cdot x - \\var{a+b*c} \\right) }\\\\
&= \\frac{{\\var{-a+b*c}} \\cdot {\\left( \\var{2*c} \\cdot x - \\var{a+b*c} \\right) }}{ (\\var{c} \\cdot x-\\var{a})^2 \\cdot (x+\\var{b})^2}
\\end{align}
\\]

d) On each subinterval\$ f'(x) \$ {ordeafg1} \$ 0 \$, which implies \$ f \$ {stijgafg1} on both \$ I_- \$ and \$ I_+ \$.

e) Since\\[   \\simplify[all,fractionNumbers]{{nulp1}} < \\simplify[all,fractionNumbers]{{a+b*c}/{2*c}} < \\simplify[all,fractionNumbers]{{nulp2}} \\]
we find
\\begin{array}{ c | c c c c c c c}
x   &            & \\simplify[all,fractionNumbers]{{nulp1}} & & \\simplify[all,fractionNumbers]{{a+b*c}/{2*c}} & & \\simplify[all,fractionNumbers]{{nulp2}} &    \\\\
    \\hline
f''(x) & \\var{{opI1tekenafg2}} & | & // & // & // & | & \\var{{opI2tekenafg2}}
\\end{array}
Consequently the curvature of \$ f\$ can be determined from
\\begin{array}{ c | c c c c c c c}
x   &            & \\simplify[all,fractionNumbers]{{nulp1}} & & \\simplify[all,fractionNumbers]{{a+b*c}/{2*c}} & & \\simplify[all,fractionNumbers]{{nulp2}} &    \\\\
    \\hline
f''(x) & \\var{{opI1tekenafg2}} & | & // & // & // & | & \\var{{opI2tekenafg2}}    \\\\
    \\hline
f'(x) & \\var{{tekenafg1}} & | & // & // & // & | & \\var{{tekenafg1}}
\\end{array}

{GetekendeGrafiek()}

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The domain of \$ f \$ is given by:

\$ \\left ] - \\infty , \\right. \$ [[0]] \$ \\left. \\right[ \\quad \\cup \\quad \\left] \\right. \$ [[1]] \$ \\left. \\, , +\\infty \\right[ \$

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What is the first order derivative of \$ f \$?
Write down your answer as a fraction with nominator an integer, and denominator in the form \$ \\left( NaturalNumber1 * x - NaturalNumber2 \\right) \\left( x - NaturalNumber3\\right) \$:

\$ f' (x) = \$ [[0]]

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Calculate the second order derivative of \$ f \$.
Write down \$ f''(x) \$ as a fraction with nominator an integer multiplied by a first order polynomial in \$ x \$ , and denominator in the form \$ \\left( NaturalNumber1 * x - NaturalNumber2 \\right)^2 \\left( x - NatuurlijkGetal3\\right)^2 \$:

\$ f'' (x) = \$ [[0]]

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The domain of \$ f \$ is the union of two open intervals, where the first of these interals, called \$I_- \$, has as endpoint \$ -\\infty \$ and the second one, \$I_+ \$, has endpoint \$ +\\infty \$.
Which of the following is correct?

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Select the correct expression concerning the curvature of \$ f \$:

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", "

"], "matrix": [0, 0, "0", 0, "if(d>0,1,0)", 0, "0", "if(d<0,1,0)"], "distractors": ["", "", "", "", "", "0", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question", "contributors": [{"name": "Paul Verheyen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3610/"}]}]}], "contributors": [{"name": "Paul Verheyen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3610/"}]}

Consider the function \\( f: \\mathbb{R} \\to \\mathbb{R} : x \\mapsto \\ln{ \\frac{\\var{c} \\cdot x-\\var{a}}{x-\\var{b}} }\\, . \\)