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{"name": "Differentiation: Stationary Points and their Nature 3", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Differentiation: Stationary Points and their Nature 3", "tags": [], "metadata": {"description": "<p>Calculating the derivative of a function of the form $\\frac{(x-a)^2}{bx}$, finding its stationary points and determining their nature.</p>", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "<p></p>\n<p></p>\n<p></p>", "advice": "<p>Part (a):</p>\n<p>Since the function is of the form $y=\\tfrac{u(x)}{v(x)}$,&nbsp;we want to use the quotient rule to calculate the derivative $\\tfrac{dy}{dx}$.</p>\n<p>As \\[ \\simplify{y=(x-{a})^2/({b}x)}, \\]</p>\n<p>let \\[ u(x) = \\simplify{(x-{a})^2} \\quad \\text{and} \\quad v(x)=\\simplify{{b}x}.\\]</p>\n<p>Next, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:</p>\n<p>\\[ \\dfrac{du}{dx} = \\simplify{2(x-{a})}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{b}}.\\]</p>\n<p>Substituting these results into the quotient rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:</p>\n<p>\\[ \\begin{split} \\dfrac{dy}{dx} &amp;\\,= \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2} \\\\ \\\\&amp;\\,=\\dfrac{(\\simplify{{b}x}) \\times\\simplify{2(x-{a})} - \\simplify{(x-{a})^2} \\times \\simplify{{b}}}{\\simplify{({b}x)^2}}.&nbsp; \\end{split}\\]</p>\n<p>Simplifying,</p>\n<p>\\[ \\begin{split} \\dfrac{dy}{dx} &amp;\\,=\\dfrac{\\simplify{2{b}x(x-{a})-{b}(x-{a})^2}}{\\simplify{{b}^2x^2}} \\\\ &amp;\\,= \\dfrac{\\simplify{2x(x-{a})-(x-{a})^2}}{\\simplify{{b}x^2}}\\\\ &amp;\\,=\\dfrac{\\simplify[!cancelTerms]{2x^2-{2a}x-x^2+{2a}x-{a^2}}}{\\simplify{{b}x^2}}\\\\ &amp;\\,=\\dfrac{\\simplify{x^2-{a^2}}}{\\simplify{{b}x^2}} \\end{split} \\]</p>\n<p></p>\n<p></p>\n<p>Part (b):</p>\n<p>A graph can help. <em>You can pan and zoom this image if you would like to.</em></p>\n<p>{geogebra_applet('https://www.geogebra.org/m/t6vskwck',[a:{a}, b:{b}])}</p>\n<p>To find the stationary points of the function, we must solve $\\tfrac{dy}{dx}=0$ for $x$. Setting $\\tfrac{dy}{dx}=0$:</p>\n<p>\\[ \\begin{split} \\dfrac{\\simplify{x^2-{a^2}}}{\\simplify{{b}x^2}}&amp;\\,= 0&nbsp; \\\\ \\implies \\simplify{x^2-{a^2}}&amp;\\,=0 \\end{split} \\]</p>\n<p>Therefore, the stationary points are when $x=\\var{abs(a)}$ and $x=\\var{-(abs(a))}$.</p>\n<p>We find the corresponding $y$-values of the stationary points by plugging these $x$-values into the initial equation:</p>\n<p>When $x=\\var{abs(a)}$,</p>\n<p>\\[\\begin{split} y &amp;\\,=\\dfrac{\\simplify[unitFactor,!cancelTerms,fractionNumbers]{({abs(a)}-{a})^2}}{\\simplify[alwaysTimes,unitFactor]{{b}{abs(a)}}} \\\\ &amp;\\,= \\simplify[fractionNumbers]{{(abs(a)-a)^2/(b*abs(a))}} .\\end{split}\\]</p>\n<p>When $x=\\var{-abs(a)}$,</p>\n<p>\\[\\begin{split} y &amp;\\,=\\dfrac{\\simplify[unitFactor,!cancelTerms,fractionNumbers]{({-abs(a)}-{a})^2}}{\\simplify[alwaysTimes,unitFactor]{{b}({-abs(a)})}} \\\\ &amp;\\,= \\simplify[fractionNumbers]{{(-abs(a)-a)^2/(b*-abs(a))}} .\\end{split}\\]</p>\n<p>Therefore, the stationary points are \\[ \\left(\\var{abs(a)},\\simplify[fractionNumbers]{{(abs(a)-a)^2/(b*abs(a))}}\\right)\\quad \\text{and} \\quad \\left(\\var{-abs(a)},\\simplify[fractionNumbers]{{(-abs(a)-a)^2/(b*(-abs(a)))}}\\right).\\]</p>\n<p>Finally, we need to determine the nature of the stationary points. To do this we want to calculate the second derivative of the initial function and then evaluate it for each $x$-value of the stationary points.&nbsp;</p>\n<p>Recall:</p>\n<ul>\n<li>If $\\tfrac{d^2y}{dx^2}&lt;0$ the stationary point is a <strong>maximum</strong>;</li>\n<li>If $\\tfrac{d^2y}{dx^2}&gt;0$ the stationary point is a&nbsp;<strong>minimum</strong>;</li>\n<li>If $\\tfrac{d^2y}{dx^2}=0$ the stationary point is a <strong>point of inflection</strong>.</li>\n</ul>\n<p>To calculate&nbsp;$\\tfrac{d^2y}{dx^2}$, we want to differentiate&nbsp;$\\tfrac{dy}{dx}$ again with respect to $x$:</p>\n<p>\\[ \\begin{split} &amp;\\frac{dy}{dx} &amp;=\\dfrac{\\simplify{x^2-{a^2}}}{\\simplify{{b}x^2}} \\\\ \\\\\\implies &amp;\\frac{d^2y}{dx^2} &amp;\\,= \\dfrac{\\simplify{({b}x^2)(2x)-{2b}x(x^2-{a^2})}}{\\simplify{({b}x^2)^2}} \\\\ &amp;&amp;\\,=\\dfrac{\\simplify[!cancelTerms]{2x^3-2x^3+{2a^2}x}}{\\simplify{{b}x^4}} \\\\ &amp;&amp;\\,=\\simplify[unitFactor]{{2a^2/simp}/({b/simp}x^3)}\\end{split}\\]</p>\n<p>For $\\left(\\var{abs(a)},\\simplify[fractionNumbers]{{(abs(a)-a)^2/(b*abs(a))}}\\right)$,&nbsp; $\\tfrac{d^2y}{dx^2} =\\simplify[fractionNumbers]{{2a^2/(b*(abs(a))^3)}}$, so it is a {maxmin1};</p>\n<p>For $\\left(\\var{-abs(a)},\\simplify[fractionNumbers]{{(-abs(a)-a)^2/(b*-abs(a))}}\\right)$,&nbsp; $\\tfrac{d^2y}{dx^2} =\\simplify[fractionNumbers]{{2a^2/(b*(-abs(a))^3)}}$, so it is a {maxmin2}.</p>\n<p></p>", "rulesets": {}, "extensions": ["geogebra"], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(-4..4 except 0)", "description": "", "templateType": "anything", "can_override": false}, "sol": {"name": "sol", "group": "Ungrouped variables", "definition": "(2a/5)^2((2a/5)-a)^3", "description": "", "templateType": "anything", "can_override": false}, "soldp": {"name": "soldp", 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