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{"name": "CB5 - Finding turning points", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "CB5 - Finding turning points", "tags": [], "metadata": {"description": "<p>Finding the stationary points of a cubic equation and determining their nature.</p>", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "<p>Given the function \\[ \\simplify{y={a}x^3+{b}x^2+{c}x+{d}} ,\\] find its stationary points and determine their nature.</p>", "advice": "<p>To find the stationary points of the function, we must solve $\\tfrac{dy}{dx}=0$ for $x$. For the function $\\simplify{y={a}x^3+{b}x^2+{c}x+{d}}$,&nbsp;</p>\n<p>\\[ \\frac{dy}{dx} = \\simplify{{3a}x^2+{2b}x+{c}}. \\]</p>\n<p>Setting $\\frac{dy}{dx}=0$ and solving for $x$:</p>\n<p>\\[ \\simplify{{3a}x^2+{2b}x+{c}} =0&nbsp; \\\\ \\\\ \\implies x=\\var{solx1dp} \\var{x1} \\text{ and } x=\\var{solx2dp} \\var{x2}. \\]</p>\n<p>Hence, the function has two stationary points at $x=\\var{solx1dp}$ and $x=\\var{solx2dp}$. To find the corresponding $y$-coordinates, we want to plug these values back into the initial equation.</p>\n<p>When $x=\\var{solx1dp}$,</p>\n<p>\\[ \\begin{split} y &amp;\\,= \\simplify[unitFactor,!cancelTerms]{{a}*({solx1dp})^3+{b}*({solx1dp})^2+{c}*({solx1dp})+{d}} \\\\ &amp;\\,=\\simplify{{soly1dp}} \\var{y1}. \\end{split} \\]</p>\n<p>When $x=\\var{solx2dp}$,&nbsp;</p>\n<p>\\[ \\begin{split} y &amp;\\,= \\simplify[unitFactor,!cancelTerms]{{a}*({solx2dp})^3+{b}*({solx2dp})^2+{c}*({solx2dp})+{d}} \\\\ &amp;\\,=\\simplify{{soly2dp}} \\var{y2}. \\end{split} \\]</p>\n<p>Therefore, the stationary points of $y=\\simplify{{a}x^3+{b}x^2+{c}x+{d}}$ are</p>\n<p>\\[ (\\simplify{{solx1dp}},\\, \\simplify{{soly1dp}}) \\, , \\,(\\simplify{{solx2dp}},\\, \\simplify{{soly2dp}}). \\]</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} = \\simplify{{3a}x^2+{2b}x+{c}}, \\\\ \\\\\\implies &amp;\\frac{d^2y}{dx^2} = \\simplify{{6a}x+{2b}}. \\end{split}\\]</p>\n<p>For $(\\simplify{{solx1dp}},\\, \\simplify{{soly1dp}})$, $\\frac{d^2y}{dx^2} = \\simplify{{check}}$, so it is a minimum.</p>\n<p>For $(\\simplify{{solx2dp}},\\, \\simplify{{soly2dp}})$, $\\frac{d^2y}{dx^2} = \\simplify{{check2}}$, so it is a maximum.</p>\n<p></p>\n<p><a href=\"https://www.sheffield.ac.uk/mash/maths-resources/first-steps#Diff\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">Use this link to find some resources which will help you revise this topic.</span></a></p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(1..3)", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(-5..5)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(-7..7)", "description": "", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "Ungrouped variables", "definition": "random(-5..5 except b)", "description": "", "templateType": "anything", 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