# Inverse function help

Tags:
1. Aug 19, 2016

### Mark53

1. The problem statement, all variables and given/known data

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].

3. 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

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

Is this correct?

2. Aug 20, 2016

### haruspex

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?

3. Aug 20, 2016

### Mark53

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

4. Aug 20, 2016

### haruspex

Hmm... I don't still. I get 1/2 and -1/2 now.

Sure, but you were just guessing, yes? How would you justify it to yourself?

5. Aug 20, 2016

### Mark53

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

6. Aug 20, 2016

### haruspex

-(3-sqrt(-8y+17))/4 with y=-1:
-(3-sqrt(+8+17))/4
-(3-sqrt(25))/4
-(3-5)/4
-(2)/4

7. Aug 20, 2016

### Mark53

my bad I was entering it in my calculator wrong
so the answer should be -(3+sqrt(-8y+17))/4

8. Aug 20, 2016

### haruspex

Yes. (You needed a calculator for that?)