For the equation y= x2+2 x2−1 , express x in terms of y and determine the values of y for which x is real.

y can take on values in:

Step-by-step explanation:

Hopefully this is the right equation:

.

Multiply both sides by :

Distribute:

Put x terms on one side together and non-x terms on opposite side.

Add y on both sides

and subtract x^2 on both sides:

Factor:

Divide both sides by (y-1):

Now square root both sides:

y cannot be 1 because we don't want to divide by 0.

Now we also need to find when (y+2)/(y-1) is positive or zero because we can't square root negative values if we intend to have just real solutions.

The fraction is zero when y=-2 because that will give us 0 on top.

I'm going to draw a number line and test which intervals gives us positive values for our expression under the square root.

--------(-2)-------(1)----------

Plugging in -3 gives us (-3+2)/(-3-1)=(-1)/(-4)=1/4 so anything before -2 works.

Let's try 0: (0+2)/(0-1)=-2 so nothing between -2 and 1 will work.

Let's try it for 2: (2+2)/(2-1)=4/1=4 so anything after 1 will work.

So y can take on values in:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!