Answer:
[tex] f^{-1}(x) = \dfrac{3}{4} \pm \dfrac{1}{4}\sqrt{8x + 9} [/tex]
Step-by-step explanation:
[tex] f^{-1}(x) = 2x^2 - 3x [/tex]
Change function notation to y.
[tex] y = 2x^2 - 3x [/tex]
Switch x and y.
[tex] x = 2y^2 - 3y [/tex]
Solve for y.
[tex] 2y^2 - 3y = x [/tex]
Complete the square on the left side. We must divide both sides by 2 to have y^2 as the leading term on the left side.
[tex] y^2 - \dfrac{3}{2}y = \dfrac{x}{2} [/tex]
1/2 of 3/2 is 3/4. Square 3/4 to get 9/16.
Add 9/16 to both sides to complete the square.
[tex] y^2 - \dfrac{3}{2}y + \dfrac{9}{16} = \dfrac{x}{2} + \dfrac{9}{16} [/tex]
Find common denominator on right side.
[tex] (y - \dfrac{3}{4})^2 = \dfrac{8x}{16} + \dfrac{9}{16} [/tex]
If X^2 = k, then [tex] X = \pm \sqrt{k} [/tex]
[tex] y - \dfrac{3}{4} = \pm \sqrt{\dfrac{1}{16}(8x + 9)} [/tex]
Simplify.
[tex] y = \dfrac{3}{4} \pm \dfrac{1}{4}\sqrt{8x + 9} [/tex]
Back to function notation.
[tex] f^{-1}(x) = \dfrac{3}{4} \pm \dfrac{1}{4}\sqrt{8x + 9} [/tex]
