Domain, Range & Inverse of a Function

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SUMMARY

This discussion focuses on determining the domain, range, and inverse of the function defined by the equation y = 8x - x². The original function is transformed into its general form, revealing its vertex at (4, 16). The domain of the original function is established as x ≤ 4, leading to two distinct inverse functions: f-1(x) = 4 - √(x - 16) for x ≥ 16 and f-1(x) = 4 + √(x - 16) for x ≤ 16. This method effectively demonstrates how to derive inverses by separating the function at its vertex.

PREREQUISITES
  • Understanding of quadratic functions and their properties
  • Knowledge of the general form of a quadratic equation
  • Familiarity with the concept of inverse functions
  • Ability to complete the square in algebraic expressions
NEXT STEPS
  • Study the properties of quadratic functions and their graphs
  • Learn how to complete the square for various quadratic equations
  • Explore the concept of one-to-one functions and their inverses
  • Investigate the implications of domain and range in inverse functions
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Students studying algebra, educators teaching quadratic functions, and anyone interested in understanding the relationship between functions and their inverses.

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Homework Statement


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How to solve part (iv) & (v)

Homework Equations


general form : [itex]y = a(x-h)^2 + k[/itex]

The Attempt at a Solution


In part (iv) for finding domain and range I converted g(x) in general form and then compared it with general form.
 
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The inverse relation of y=8x-x2 is x=8y-y2.

Also, the domain of the original relation is the range of the inverse relation in general. Specifics depend...
 
I don't believe that is quite what is being asked. Completing the square, 8x- x2= 16- 16+ 8x- x2= 16- (x-4)2. The graph of that is a parabola having vertex at (4, 16). The function f(x)= 16- (x- 4)2 with [itex]x\le 4[/itex] has inverse function [itex]f^-1(x)= 4- \sqrt{x- 16}[/itex] while the function f(x)= 16- (x- 4)2 with [itex]x\ge 4[/itex][itex]has inverse function [itex]f^-1(x)= 4+ \sqrt{x- 16}[/itex].<br /> <br /> By separating at the vertex, we cut the given function into to one-to-one that now have inverses.[/itex]
 

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