SOLUTION: Hi I asked a question in this link but the person got it wrong, can someone try again please? Thank you. https://www.algebra.com/tutors/students/your-answer.mpl?question=1073789

Algebra ->  Inverses -> SOLUTION: Hi I asked a question in this link but the person got it wrong, can someone try again please? Thank you. https://www.algebra.com/tutors/students/your-answer.mpl?question=1073789      Log On


   

Question 1073969: Hi I asked a question in this link but the person got it wrong, can someone try again please? Thank you.
https://www.algebra.com/tutors/students/your-answer.mpl?question=1073789
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
We sometimes use symbols that are difficult to interpret, or make mistakes while typing (or while thinking), but we try.

The function f%28x%29+=+2x%5E2-4x-5 , with all real numbers as its domain, graph%28300%2C300%2C-4%2C6%2C-10%2C10%2C2x%5E2-4x-5%29 is not invertible,
because (like all quadratic functions) it has a vertical axis of symmetry.
That axis of symmetry is the line x=1 : .
By restricting the domain of the function
to a domain that does not include points on both sides of that axis,
we can make f%28x%29+=+2x%5E2-4x-5 invertible.
Defining the function only for a set of points with x%3E=1 ,
or to a set of points with x%3C=1 , makes the function invertible.
If we want to include the point x=0 in the domain,
we cannot include any point with x%3E1 ,
but we can include any point with x%3C=1 .
The largest restricted domain that includes the point x=0 ,
but does not contain any point with x%3E1 is
all real x such that x%3C=1 .
That can be expressed in interval notation as
%22%28%22-infinity%22%2C+1+%5D%22 or highlight%28%22%28+-infinity+%2C+1+%5D%22%29.
That is the largest interval that includes the point x=0,
where we can make f%28x%29+=+2x%5E2-4x-5 invertible.
That is the largest such interval that includes the point x=0.

The equation for the axis of symmetry can be found by transforming the function, or by applying a formula.
With formulas:
All quadratic functions can be written in the form y=ax%5E2%2Bbx%2Bc ,
for some constants a%3C%3E0 , b and c .
The axis of symmetry for such a function is the line x=%28-b%29%2F%222+a%22
f%28x%29+=+2x%5E2-4x-5 is a quadratic function with a=2 and b=-4 ,
so its axis of symmetry is
x=%28-%28-4%29%29%2F%282%2A2%29 --> x=4%2F4 ---> x=1 .
Transforming the function:
f%28x%29+=+2x%5E2-4x-5 or y+=+2x%5E2-4x-5
y=2x%5E2-4x%2B2-7
y=2%28x%5E2-2x%2B1%29-7
y=2%28x-1%29%5E2-7

The inverse for y=2%28x-1%29%5E2-7 forall x%3C=1 ,graph%28300%2C300%2C-4%2C6%2C-10%2C10%2C2x%5E2-4x-5%2Bsqrt%281-x%29-sqrt%281-x%29%29 is
system%28x=2%28y-1%29%5E2-7%2C%22with%22%2Cy%3C=1%29 ---> system%28%28x%2B7%29%2F2=%28y-1%29%5E2%2C%22with%22%2Cy-1%3C=0%29 ---> y-1=-sqrt%28%28x%2B7%29%2F2%29 ---> y=1-sqrt%28%28x%2B7%29%2F2%29 ,graph%28300%2C300%2C-10%2C10%2C-4%2C6%2C1-sqrt%28%28x%2B7%29%2F2%29%29
or f%5E%28-1%29%28x%29=1-sqrt%28%28x%2B7%29%2F2%29 .

Here are the graphs for blue%28f%28x%29%29, red%28f%5E%28-1%29%29, and green%28y=x%29 , which is the line you flip bule%28f%29 over to get red%28f%5E%28-1%29%29 :