Proving if a function is One to One

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SUMMARY

The discussion focuses on determining whether specific functions are one-to-one and finding their inverses if applicable. For the function a) f(x) = cos x, defined on the interval x ∈ [-π/2, π/2], it is concluded that it is not one-to-one due to failing the horizontal line test and having equal outputs for different inputs. Function b) f(x) = x/|x| is also confirmed as not one-to-one, as it produces the same output for distinct inputs, specifically f(1) = f(2). Lastly, function c) f(x) = (2 - 3x²)³ is determined to be not one-to-one as it has multiple roots, evidenced by f(√(2/3)) = f(-√(2/3)).

PREREQUISITES
  • Understanding of one-to-one functions and the horizontal line test
  • Knowledge of derivatives and their implications on function behavior
  • Familiarity with trigonometric functions, specifically the cosine function
  • Basic algebraic manipulation involving absolute values and roots
NEXT STEPS
  • Study the properties of one-to-one functions in detail
  • Learn about the horizontal line test and its applications
  • Explore the implications of derivatives on function monotonicity
  • Investigate the behavior of piecewise functions and their inverses
USEFUL FOR

Students studying calculus, particularly those focusing on function analysis, derivatives, and inverse functions. This discussion is beneficial for anyone needing clarification on one-to-one functions and their characteristics.

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Homework Statement


Determine whether or not the function is one-to-one and, if so, find the inverse. If the function has an inverse, give the domain of the inverse.
a) f(x) = cos x, x[itex]\in[-pi/2, pi/2][/itex]
b) f(x) = x/|x|
c) f(x) = (2-3x2)3

Homework Equations





The Attempt at a Solution


a) f '(x) = -sinx
When graphing the original function, you can see that it does not pass the horizontal line test. The derivative shows that between this interval, the derivative is not always positive or always negative, it changes between the two which proves that the original function increases and decreases between this interval. Also, f(-pi/3) = f(pi/3) because cos (-pi/3) = 1/2 = cos(pi/3) and that proves that this function is not one-to-one.

b) This is the question I'm most confused about between the three because when i graphed it using wolfram alpha, it is clear that it is not a one to one function because the graph consists of a horizontal line of y = -1 and y = 1 and that clearly will not pass the horizontal line test. However, i just don't know how to show that this function is not one-to-one. This is what i think using what i did before:
1/|1| = 1 = 2/|2| which proves that f(1) = f(2) and that means that this function is not one-to-one

c) I took the derivative of the function:
f'(x) = -18x(2-3x2)2
From the derivative you can see that the function is not always increasing or always decreasing, it changes between the two depending on what the value of x is. I can tell by looking at the function that it has more than one root and if you can show that the function has more than one root that automatically proves that the function is not one-to-one because it will not pass the horizontal line test.

(2-3x2)3 = 0
(2-3x2) = 0
3x2 = 2
x = +-√(2/3)
that means that f(+√(2/3))= f(-√(2/3)) which proves that the function is not one-to-one
Please help me verify if these are correct, and specific help on part b with the absolute value will be greatly appreciated. Thank you.
 
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(nevermind, I see that you are using the definition of one-to-one)

Part a looks good. You have [itex]cos(-\frac{\pi}{3}) = cos(\frac{\pi}{3})[/itex] but [itex]-\frac{\pi}{3} \neq \frac{\pi}{3}[/itex].

Part b also looks good. You have found that f(1) == f(2) but obviously 1 is not equal to 2.

Part c also looks good.
 
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