Finding the Inverse Function of f(x) = 1−3x−2x^2 on Domain [-2, -1]

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SUMMARY

The discussion focuses on finding the inverse function of f(x) = 1 - 3x - 2x² over the domain [-2, -1]. Participants confirm that the function is one-to-one using the Horizontal Line Test and establish the range R as [-1, 2]. The correct expression for the inverse function is derived as f⁻¹(y) = -(3 + √(-8y + 17))/4, with emphasis on verifying calculations to avoid sign errors and misinterpretations.

PREREQUISITES
  • Understanding of the Horizontal Line Test for one-to-one functions
  • Knowledge of quadratic equations and the quadratic formula
  • Familiarity with function notation and inverse functions
  • Basic algebraic manipulation and error checking techniques
NEXT STEPS
  • Study the application of the Horizontal Line Test in various functions
  • Learn about the properties of inverse functions and their derivations
  • Explore quadratic functions and their ranges in detail
  • Practice error-checking techniques in algebraic calculations
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Students studying calculus, particularly those focusing on functions and their inverses, as well as educators looking for examples of function analysis and error correction in mathematical computations.

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Homework Statement



Let f(x) = 1−3x−2x^2 , x ∈ [−2, −1]. Use the Horizontal Line Test to show that f is 1–1 (on its given domain), and find the range R of f. Then find an expression for the inverse function f −1 : R → [−2, −1].

The Attempt at a Solution


I have already done the horizontal line test but I am unsure about my working out for the other parts below

would the range just be:

f(-2)=-1
f(-1)=2

y ∈ [−1, 2]

finding expression for inverse function

1−3x−2x^2=y
-2x^2-3x-y+1=0
using quadratic formula

x=(3-sqrt(-8y+17)/4 as (3+sqrt(-8y+17)/4 lies outside the range

Is this correct?
 
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Mark53 said:
x=(3-sqrt(-8y+17)/4
Sign error, and missing parenthesis. You should have checked whether it was right by substituting the values of x.
Mark53 said:
would the range just be:
Yes, but how do you justify it?
Note that f is continuous and 1-1. What does that tell you about turning points?
 
haruspex said:
Sign error, and missing parenthesis. You should have checked whether it was right by substituting the values of x.

Yes, but how do you justify it?
Note that f is continuous and 1-1. What does that tell you about turning points?

x=-(3-sqrt(-8y+17))/4

when I sub in y=-1 and 2 I get the correct x values

Is this correct now?

The question doesn't say anything about justifying the range though
 
Mark53 said:
x=-(3-sqrt(-8y+17))/4
when I sub in y=-1 and 2 I get the correct x values
Hmm... I don't still. I get 1/2 and -1/2 now.

Mark53 said:
The question doesn't say anything about justifying the range
Sure, but you were just guessing, yes? How would you justify it to yourself?
 
haruspex said:
Hmm... I don't still. I get 1/2 and -1/2 now.

I get a 1/2 and -1/2 when x=-(3+sqrt(-8y+17))/4

but when x=-(3-sqrt(-8y+17))/4
i get -1 and 2
 
Mark53 said:
I get a 1/2 and -1/2 when x=-(3+sqrt(-8y+17))/4

but when x=-(3-sqrt(-8y+17))/4
i get -1 and 2
-(3-sqrt(-8y+17))/4 with y=-1:
-(3-sqrt(+8+17))/4
-(3-sqrt(25))/4
-(3-5)/4
-(2)/4
 
haruspex said:
-(3-sqrt(-8y+17))/4 with y=-1:
-(3-sqrt(+8+17))/4
-(3-sqrt(25))/4
-(3-5)/4
-(2)/4
my bad I was entering it in my calculator wrong
so the answer should be -(3+sqrt(-8y+17))/4
 
Mark53 said:
my bad I was entering it in my calculator wrong
so the answer should be -(3+sqrt(-8y+17))/4
Yes. (You needed a calculator for that?)
 
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