Find the domain and range of a real function f(x) = (x^2-1)/(x+1)?

Question

lgbasallote · Answer

you have $$\frac{x^2 - 1}{x+1}$$
the domain restriction will be the value of x that will make x + 1 = 0

mathteacher1729 · Answer

Domain: 
zero in denominator? 
log of negative number or zero?
sqrt of a negative number?

Range: 
Graphing  or
Finding domain of inverse  or
arguments from the form of the function itself (a bit more advanced)

anonymous · Answer

Domain = R-[-1]

anonymous · Answer

Range=??

anonymous · Answer

see x!=-1 that's domain  and range is x!=-2

mathteacher1729 · Answer

The range is all values of y that the function can take on. 

$$y = \frac{(x-1)(x+1)}{(x+1)}$$ we can cancel and write 

y = x - 1

This has range of all real numbers. 

BUT 

our original function did not allow for x = -1 which means y cannot equal (-1 - 1) = -2. 

So the range is all real numbers except y = -2.

anonymous · Answer

but the answer in my book says range = R

mathteacher1729 · Answer

That is incorrect. If the range was all real number , that means we can solve for ANY real number we choose. 

Solve 

$$\huge -2 = \frac{x^2-1}{x+1}$$

You cannot do this. You would need to have x = -1 after canceling common factors, and ... x cannot equal -1.

mathteacher1729 · Answer

See attached.

anonymous · Answer

I m on iPad now and this is not allowin me to draw or write equation it is also getting unstable sorry can't help write now

anonymous · Answer

@satellite73  plzz  help

anonymous · Answer

i am confused plzz help

anonymous · Answer

the answer in the book is wrong as mathteacher said
pity

anonymous · Answer

Can u give the explanation

anonymous · Answer

is the Domain = R-[-1]???

anonymous · Answer

Agreed even I think book is wrong

anonymous · Answer

$$f(x)=\frac{x^2-1}{x+1}=\frac{(x+1)(x-1)}{x+1}=x-1$$ only provided $x\neq -1$

anonymous · Answer

then R=??

anonymous · Answer

if $x=-1$ you would get 
$$f(-1)=\frac{(-1)^2-1}{-1+1}=\frac{1-1}{-1+1}=\frac{0}{0}$$ and although you can write it with pencil and paper, $\frac{0}{0}$ is not a number
that is why the domain excludes $-1$ because you cannot divide by zero

anonymous · Answer

ok got it  then range=?

anonymous · Answer

therefore the domain is "all real numbers except  -1" which can be written as 
$$\mathbb{R}-[-1]$$ if you like

anonymous · Answer

Then Range=?

anonymous · Answer

now for the range
 $f(x)=x-1$ for all $x$ except $-1$ where the function is undefined
$y=x-1$ is a line with slope 1 and $y$ intercept $-1$ which is exactly what the graph looks like
therefore the range is all real numbers, except the number you would get if you replace $x$ by $-1$ because $-1$ is not in the domain 
if you replace $x$ by $-1$ you get 
$$f(-1)=-1-1=-2$$ and so this function can never be $-2$ because you may not replace $x$ by $-1$

anonymous · Answer

so Range= R-[-2]

anonymous · Answer

another way to say this is that there is no input for the function 
$$f(x)=\frac{x^2-1}{x+1}$$ that would give the answer $-2$ 
if you want we could try and solve it, and see why it doesn't work
yes, it is what you wrote

anonymous · Answer

thxxx

anonymous · Answer

yw
hope it is clear

anonymous · Answer

Sorry cudnt help yahoo