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

AI Thread Summary
The discussion focuses on finding the inverse function of f(x) = 1−3x−2x^2 over the domain [-2, -1]. Participants confirm that the function is one-to-one using the Horizontal Line Test and discuss the range, concluding it is [-1, 2]. The inverse function is derived as x = -(3 + sqrt(-8y + 17))/4, with corrections made for sign errors and proper substitution checks. There is some confusion about justifying the range, but the final expression for the inverse function is validated through calculations. Overall, the thread emphasizes the importance of careful computation and verification in solving for inverse functions.
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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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