Answer:
{xcR} (x is all real numbers) and {y[tex]\leq[/tex]2[tex]\sqrt{\frac{y-1}{2} }[/tex]+1}
Step-by-step explanation:
1st: Find x by squaring both sides, subtracting 1 from both sides, and dividing by 2 from both sides to get x= [tex]\sqrt{\frac{y-1}{2} }[/tex]
2nd: Plug in x for y=[tex]2x^{2}[/tex]+1 which equals y=2([tex]\sqrt{\frac{y-1}{2} }[/tex])+1
3rd: This gives you the range (y) which is {y[tex]\leq[/tex]2[tex]\sqrt{\frac{y-1}{2} }[/tex]+1}
4th: domain is all real numbers
I hope this helped because this took me 30 minutes to do. I had to pull out my algebra math book to check I did it right. I hope I was right!
