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Let $y = \\simplify{{a1}x^2 + {b1}x + {c1}}$.

\n

$\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} = $ [[0]]

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Enter the coordinates of the stationary point of $y$: $\\big($ [[1]] $, $ [[2]] $\\big)$

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$\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}x^2} = $ [[3]]

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What is the nature of the stationary point of $y$?

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[[4]]

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Maximum

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Point of inflection

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Let $z = \\simplify{{a2}x^2+{b2}x+{c2}}$.

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$\\dfrac{\\mathrm{d}z}{\\mathrm{d}x} = $ [[0]]

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Enter the coordinates of the stationary point of $z$: $\\big($ [[1]] $, $ [[2]] $\\big)$

\n

$\\dfrac{\\mathrm{d}^2z}{\\mathrm{d}x^2} = $ [[3]]

\n

What is the nature of the stationary point of $z$?

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[[4]]

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Point of inflection

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Let $t = \\var{m}x^3+\\var{q}$.

\n

$\\dfrac{\\mathrm{d}t}{\\mathrm{d}x} = $ [[0]]

\n

Enter the coordinates of the stationary point of $t$: $\\big($ [[1]] $, $ [[2]] $\\big)$

\n

$\\dfrac{\\mathrm{d}^2t}{\\mathrm{d}x^2} = $ [[3]]

\n

What is the nature of the stationary point of $t$?

\n

[[4]]

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a)

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$\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} = \\simplify{{2a1}x+{b1}}$.

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To find the $x$-coordinate of the stationary point, solve $\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 0$ for $x$:

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\\[ \\begin{align} \\frac{\\mathrm{d}y}{\\mathrm{d}x} = \\simplify{{2a1}x + {b1}} &= 0 \\\\ x &= \\simplify[all,fractionNumbers]{{sx1}} \\end{align} \\]

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Find the $y$-coordinate by substituting this value of $x$ into the definition of $y(x)$:

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\\[\\begin{align} \\simplify[fractionnumbers]{y({sx1})} &= \\simplify[basic,fractionnumbers]{{a1}{sx1}^2+{b1}{sx1}+{c1}} \\\\ &= \\simplify[fractionnumbers]{{sy1}} \\end{align}\\]

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Finally, to determine the nature of the stationary point, look at $\\frac{\\mathrm{d}^2y}{\\mathrm{d}x^2}$ at $x = \\simplify[fractionnumbers]{{sx1}}$.

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\\[ \\frac{\\mathrm{d}^2y}{\\mathrm{d}x^2} = \\simplify{{2*a1}} \\]

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This is positive, so the stationary point is a minimum.

\n

b)

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$\\dfrac{\\mathrm{d}z}{\\mathrm{d}x} = \\simplify{{2a2}x+{b2}}$.

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To find the $x$-coordinate of the stationary point, solve $\\frac{\\mathrm{d}z}{\\mathrm{d}x} = 0$ for $x$:

\n

\\[ \\begin{align} \\frac{\\mathrm{d}z}{\\mathrm{d}x} = \\simplify{{2a2}x + {b2}} &= 0 \\\\ x &= \\simplify[all,fractionNumbers]{{sx2}} \\end{align} \\]

\n

Find the $y$-coordinate by substituting this value of $x$ into the definition of $z(x)$:

\n

\\[\\begin{align} \\simplify[fractionnumbers]{z({sx1})} &= \\simplify[basic,fractionnumbers]{{a2}{sx2}^2+{b2}{sx2}+{c2}} \\\\ &= \\simplify[fractionnumbers]{{sy2}} \\end{align}\\]

\n

Finally, to determine the nature of the stationary point, look at $\\frac{\\mathrm{d}^2z}{\\mathrm{d}x^2}$ at $x = \\simplify[fractionnumbers]{{sx2}}$.

\n

\\[ \\frac{\\mathrm{d}^2z}{\\mathrm{d}x^2} = \\simplify{{2*a2}} \\]

\n

This is negative, so the stationary point is a maximum.

\n

c)

\n

$\\dfrac{\\mathrm{d}t}{\\mathrm{d}x} = \\simplify{{3m}x^2}$.

\n

To find the $x$-coordinate of the stationary point, solve $\\frac{\\mathrm{d}t}{\\mathrm{d}x} = 0$ for $x$:

\n

\\[ \\begin{align} \\frac{\\mathrm{d}t}{\\mathrm{d}x} = \\simplify{{3m}x^2} &= 0 \\\\ x &= 0 \\end{align} \\]

\n

Find the $y$-coordinate by substituting this value of $x$ into the definition of $t(x)$:

\n

\\[\\begin{align} z(0) &= \\simplify[basic,fractionnumbers]{{m}*0^3 + {q}} \\\\ &= \\var{q} \\end{align}\\]

\n

Finally, to determine the nature of the stationary point, look at $\\frac{\\mathrm{d}^2t}{\\mathrm{d}x^2}$ at $x = 0$.

\n

\\[ \\frac{\\mathrm{d}^2t}{\\mathrm{d}x^2} = \\simplify{{6m}x} \\]

\n

\\[ \\left.\\frac{\\mathrm{d}^2t}{\\mathrm{d}x^2} \\right\\rvert_{x=0} = \\var{6m} \\times 0 = 0 \\]

\n

This is zero, so the stationary point is a point of inflection.

", "statement": "

For the following, find the stationary point and determine its nature. For your answers, where appropriate, write your solutions as fractions, NOT decimals, and cancel down where possible.

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