Prove that $f(x)$ surjective.
\nAssume that $b \\in \\mathbb R$. We must find some $a$ such that $f(a) = b$. There are two cases to consider.
\nIf $b=0$ then choose $a = 0$.
\nOtherwise, $b \\neq 0$ and then choose $a = $ [[0]].
\nIn either case, $f(a) = b$ and so the function is surjective.
", "extendBaseMarkingAlgorithm": true, "type": "gapfill", "sortAnswers": false, "showCorrectAnswer": true, "showFeedbackIcon": true, "unitTests": [], "marks": 0, "customMarkingAlgorithm": "", "scripts": {}, "variableReplacementStrategy": "originalfirst"}, {"variableReplacements": [], "gaps": [{"variableReplacements": [], "notationStyles": ["plain", "en", "si-en"], "extendBaseMarkingAlgorithm": true, "correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "maxValue": "0", "mustBeReducedPC": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "unitTests": [], "marks": 1, "correctAnswerStyle": "plain", "customMarkingAlgorithm": "", "allowFractions": false, "minValue": "0", "scripts": {}, "variableReplacementStrategy": "originalfirst"}], "prompt": "Prove that $f(x)$ is injective.
\nAssume that $f(x_0) = f(x_1)$. We must prove that $x_0 = x_1$. There are two cases to consider.
\nIf $f(x_0) = f(x_1) = 0$ then $x_0 = x_1 = $ [[0]].
\nOtherwise, $f(x_0) = f(x_1) \\neq 0$ and then
\n$x_0 = \\frac{1}{f(x_0)} = \\frac{1}{f(x_1)} = x_1$.
\nIn either case, $x_0 = x_1$ and so the function is injective.
", "extendBaseMarkingAlgorithm": true, "type": "gapfill", "sortAnswers": false, "showCorrectAnswer": true, "showFeedbackIcon": true, "unitTests": [], "marks": 0, "customMarkingAlgorithm": "", "scripts": {}, "variableReplacementStrategy": "originalfirst"}, {"variableReplacements": [], "gaps": [{"variableReplacements": [], "checkVariableNames": false, "vsetRange": [0, 1], "variableReplacementStrategy": "originalfirst", "extendBaseMarkingAlgorithm": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "type": "jme", "showCorrectAnswer": true, "showFeedbackIcon": true, "showPreview": true, "unitTests": [], "vsetRangePoints": 5, "marks": 1, "failureRate": 1, "customMarkingAlgorithm": "", "expectedVariableNames": [], "scripts": {}, "answer": "x"}], "prompt": "Because $f$ is surjective and injective, it is bijective and therefore has an inverse.
\nIn this case we can actually see that $f$ is its own inverse because $f(f(x)) = $ [[0]].
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\nThis is an important property because it guarantees that $f: A \\mapsto B$ has an inverse function $f^{-1} : B \\mapsto A$.
\nLet's prove that the function $f: \\mathbb R \\mapsto \\mathbb R$ defined by
\n$ f(x) = \\left\\{ \\begin{array}{cl} \\frac{1}x & x \\neq 0 \\\\ 0 & x = 0.\\end{array}\\right.$
\nhas an inverse.
\n", "advice": "This question summarizes the definitions of surjective and injective, and applies them to prove the existance of an inverse.
\nFor continuous functions it is possible to prove that it injective by using the derivative. But $f$ is not continuous, and nor is it strictly increasing or strictly decreasing. In fact it is strictly decreasing for $x<0$ and strictly increasing for $x>0$.
\n{myscript()}
\nSo we pretty much have to prove that $f$ is injective directly from the definition.
\nSince $f$ is a bijection we know that an inverse exists, but this information does not tell us how to find it. This example is special because the inverse is easy to find.
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