Question

What is the inverse and restricted domain of the equation 8x^2-3

Answer 1

Answer:  $\bold{y=\pm \dfrac{\sqrt{2(x+3)}}{4},\qquad x\neq -3}$

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

$\dfrac{x+3}{8}=y^2$           divided both sides by 8

$\sqrt{\dfrac{x+3}{8}}=\sqrt{y^2}$           square rooted both sides

$\pm \sqrt{\dfrac{x+3}{8}}=y$           simplified

$\pm \sqrt{\dfrac{x+3}{8}\bigg(\dfrac{2}{2}\bigg)}=y$           rationalized the denominator

$\pm \sqrt{\dfrac{2(x+3)}{16}}=y$           simplified

$\pm \dfrac{\sqrt{2(x+3)}}{4}=y$           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