#1**+2 **

Solve each equation for 'x' ....then switch the 'x' and 'y'....

I'll do the third one....you try the other two

y = sqrt(5x-6) square both sides

y^2 = 5x-6 add 6 to both sides

y^2 + 6 = 5x divide both sides by 5

(y^2+6)/5 = x now swithch the 'x's and 'y's

y = (x^2+6) / 5 done!

ElectricPavlov Apr 20, 2019

#2**+2 **

a) y = x^2 + 3 subtract 3 from both sides

y - 3 = x^2 take both roots

±√ [ y - 3 ] = x "swap" x and y

±√[ x - 3] = y Note, GM.....this is the inverse.....but not an inverse "function".....since the original function isn't one-to-one.....then we don't have a true inverse function

b) y = [ (1/4)x + 6 ] ^3 take the cube root of both sides

∛y = (1/4)x + 6 subtract 6 from both sides

∛[y] - 6 = (1/4) x multiply through by 4

4 ∛[y] - 24 = x swap x and y

4 ∛[x] - 24 = y = the inverse

CPhill Apr 21, 2019