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Not your average video:
Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.
Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.
Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.
Function Theory
Watch comprehensive video reviews for Function Theory, designed for final exam preparation. Each video includes integrated problems you can solve alongside detailed solutions.
Not your average video:
Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.
Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.
Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.
Not your average video:
Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.
Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.
Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.
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AHL 2.14
When a function f is not one to one, ie it does not pass the horizontal line test, its domain needs to be restricted for the inverse to exist.
Example
Let f(x)=(x−1)2+2, and g(x)=f(x) for x≥a. Find the smallest possible value of a such that g−1 exists.
The parabola (x−1)2+2 has a vertex at (1,2). We therefore need to "cut" the function there so that the inverse exists. Hence a=1.
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AHL 2.14
When a function f is not one to one, ie it does not pass the horizontal line test, its domain needs to be restricted for the inverse to exist.
Example
Let f(x)=(x−1)2+2, and g(x)=f(x) for x≥a. Find the smallest possible value of a such that g−1 exists.
The parabola (x−1)2+2 has a vertex at (1,2). We therefore need to "cut" the function there so that the inverse exists. Hence a=1.
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