// Numbas version: exam_results_page_options {"name": "Functions: Inverse Functions 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Functions: Inverse Functions 3", "tags": [], "metadata": {"description": "

Finding the inverse of a function of the form $f(x)=\\frac{x^2}{a}+b$ for $x \\geq 0$.

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If $f(x)=\\simplify{x^2/{a}+{b}}$ for $x \\geq 0$, find the inverse function, $f^{-1}(x)$.

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To find $f^{-1}x$, it can help to first set $f(x)$ to a different variable, which we will call $y$:

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\\[ y = f(x) = \\simplify{x^2/{a}+{b}}\\]

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Since the function $f(x)$ takes us from $x$ to $y$, the inverse function $f^{-1}$ will take us from $y$ to $x$. So to obtain $f^{-1}$, we want to rearrange $y=\\simplify{x^2/{a}+{b}}$ so that it is $x$ as a function of $y$:

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\\[ \\begin{split} y &\\,= \\simplify{x^2/{a}+{b}} \\\\ \\simplify{y-{b}} &\\,= \\simplify{x^2/{a}} \\\\ \\simplify{{a}(y-{b})} &\\,= x^2 \\\\ \\implies x &\\,=\\sqrt{\\simplify{{a}(y-{b})}}. \\end{split} \\]

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Hence, $f^{-1}(y) =\\sqrt{\\simplify{{a}(y-{b})}}$, and therefore \\[ f^{-1}(x) =\\sqrt{\\simplify{{a}(x-{b})}}.\\]

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(Note: The inverse is valid for all values of $x\\geq\\var{b}$.)

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$f^{-1}(x)=$[[0]]

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