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

In summary, for the given function f(x) = 1-3x-2x^2 on the domain x ∈ [-2, -1], the Horizontal Line Test shows that it is 1-1 and the range R is given by y ∈ [-1, 2]. To find the inverse function f^-1, the quadratic formula can be used to obtain x = -(3+sqrt(-8y+17))/4.
  • #1
Mark53
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0

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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  • #2
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?
 
  • #3
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
 
  • #4
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?
 
  • #5
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
 
  • #6
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
 
  • #7
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
 
  • #8
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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1. What is an inverse function?

An inverse function is a mathematical operation that undoes another function. It essentially reverses the input and output of the original function.

2. How do you find the inverse of a function?

To find the inverse of a function, you must switch the x and y variables and then solve for y. This can be done algebraically or graphically.

3. What is the domain and range of an inverse function?

The domain of an inverse function is the range of the original function, and the range of an inverse function is the domain of the original function. In other words, the input and output values are swapped.

4. How do you test if two functions are inverses of each other?

To test if two functions are inverses of each other, you can use the composition test. This involves plugging one function into the other and seeing if the result is the input value.

5. Can every function have an inverse?

No, not every function has an inverse. For a function to have an inverse, it must be one-to-one, meaning each input has a unique output. Functions that fail the horizontal line test do not have an inverse.

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