Question

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

Answer 1

Answer:
`\bold{y=\pm (√(2(x+3)))/(4),\qquad x\\eq -3}`

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

`(x+3)/(8)=y^2` divided both sides by 8

`\sqrt{(x+3)/(8)}=√(y^2)` square rooted both sides

`\pm \sqrt{(x+3)/(8)}=y` simplified

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

`\pm \sqrt{(2(x+3))/(16)}=y` simplified

`\pm (√(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