Answer :
Answer: [tex]\bold{y=\pm \dfrac{\sqrt{2(x+3)}}{4},\qquad x\neq -3}[/tex]
Step-by-step explanation:
y = 8x² - 3 (Restriction: none - x is All Real Numbers)
The inverse is when you swap the x's and y's and then solve for y
x = 8y² - 3 swapped the x and y
x + 3 = 8y² added 3 to both sides
[tex]\dfrac{x+3}{8}=y^2[/tex] divided both sides by 8
[tex]\sqrt{\dfrac{x+3}{8}}=\sqrt{y^2}[/tex] square rooted both sides
[tex]\pm \sqrt{\dfrac{x+3}{8}}=y[/tex] simplified
[tex]\pm \sqrt{\dfrac{x+3}{8}\bigg(\dfrac{2}{2}\bigg)}=y[/tex] rationalized the denominator
[tex]\pm \sqrt{\dfrac{2(x+3)}{16}}=y[/tex] simplified
[tex]\pm \dfrac{\sqrt{2(x+3)}}{4}=y[/tex] simplified
Restriction:
The radical (inside the square root sign) cannot be negative
→ 2(x + 3) ≥ 0
x + 3 ≥ 0 divided both sides by 2
x ≥ -3 subtracted 3 from both sides