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

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Homework Help Overview

The discussion revolves around finding the inverse function of f(x) = 1−3x−2x^2 on the domain [-2, -1]. Participants explore the Horizontal Line Test to establish that the function is one-to-one and attempt to determine the range of the function as well as the expression for its inverse.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the Horizontal Line Test and the continuity of the function. There are attempts to calculate the range and to derive the inverse function using the quadratic formula. Questions arise regarding the justification of the range and the correctness of the derived expressions.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts. Some guidance has been offered regarding the need for justification of the range and the correctness of the expressions derived for the inverse function. There is ongoing exploration of the calculations and potential errors in the expressions.

Contextual Notes

Participants note that the original question does not explicitly require justification for the range, leading to some uncertainty in the discussion. There are also mentions of sign errors and the need for careful substitution to verify results.

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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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