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Having been given the derivatives at a point $a = \\var{pointQ}$, we can reconstruct the cubic function using Taylor's theorem: since

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\\[\\begin{align*}p(x) &= \\sum_{i=0}^3 \\frac{p^{(i)}(a)}{i!} (x-a)^i \\\\ &= \\frac{\\var{derivs[0]}}{0!}(\\simplify{x-{pointQ}})^0 + \\frac{\\var{derivs[1]}}{1!} (\\simplify{x-{pointQ}})^1 + \\frac{\\var{derivs[2]}}{2!} (\\simplify{x-{pointQ}})^2 + \\frac{\\var{derivs[3]}}{3!} (\\simplify{x-{pointQ}})^3 \\\\ &= \\simplify[all,!collectNumbers,!noLeadingMinus]{{derivs[0]} + {derivs[1]} (x-{pointQ}) + {derivs[2]}/2 (x-{pointQ})^2 + {derivs[3]}/6 (x-{pointQ})^3},\\end{align*}\\]

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into which we can then substitute $x = \\var{pointA}$ in order to find $p(\\var{pointA}) = \\var{valueA}$.

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If you expanded the brackets to find the function in its simplest form, you should have obtained

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\\[p(x) = \\simplify[all,!collectNumbers,!noLeadingMinus]{{coeffs[0]} + {coeffs[1]} x + {coeffs[2]} x^2 + {coeffs[3]} x^3}\\]

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What is $p(\\var{pointA})$?

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I am thinking of a cubic function. Its derivatives at $a = \\var{pointQ}$ are the following:

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\\[\\begin{align*} p(\\var{pointQ}) &= \\var{derivs[0]}; \\\\ p^{(1)}(\\var{pointQ}) &= \\var{derivs[1]}; \\\\ p^{(2)}(\\var{pointQ}) &= \\var{derivs[2]}; \\text{ and} \\\\ p^{(3)}(\\var{pointQ}) &= \\var{derivs[3]}. \\end{align*}\\]

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Given the derivatives of a cubic function at a point, find the value of the function at another point.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "contributors": [{"name": "Philip Walker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/362/"}]}]}], "contributors": [{"name": "Philip Walker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/362/"}]}