Question

Find the inverse when f(x)=3x^2-3x-2

Answer 1

Answer:

The inverse of f(x) is $f^{-1}$(x) =  ± $\sqrt{\frac{x+\frac{11}{4}}{3}}$ + $\frac{1}{2}$

Step-by-step explanation:

To find the inverse of the quadratic function f(x) = ax² + bx + c, you should put it in the vertex form f(x) = a(x - h)² + k, where

• h = $\frac{-b}{2a}$

• k is the vlue f at x = h

∵ f(x) = 3x² - 3x - 2

→ Compare it with the 1st form above to find a and b

∴ a = 3 and b = -3

→ Use the rule of h to find it

∵ h = $\frac{-(-3)}{2(3)}$ = $\frac{3}{6}$ = $\frac{1}{2}$

∴ h =  $\frac{1}{2}$

→ Substitute x by the value of h in f to find k

∵ k = 3( $\frac{1}{2}$)² - 3( $\frac{1}{2}$) - 2

∴ k = $-\frac{11}{4}$

→ Substitute the values of a, h, and k in the vertex form above

∵ f(x) = 3(x -  $\frac{1}{2}$)² + $-\frac{11}{4}$

∴ f(x) = 3(x -  $\frac{1}{2}$)² - $\frac{11}{4}$

Now let us find the inverse of f(x)

∵ f(x) = y

∴ y = 3(x -  $\frac{1}{2}$)² - $\frac{11}{4}$

→ Switch x and y

∵ x = 3(y -  $\frac{1}{2}$)² - $\frac{11}{4}$

→ Add $\frac{11}{4}$ to both sides

∴ x + $\frac{11}{4}$ = 3(y - $\frac{1}{2}$)²

→ Divide both sides by 3

∵ $\frac{x+\frac{11}{4}}{3}$ = (y - $\frac{1}{2}$)²

→ Take √ for both sides

∴ ± $\sqrt{\frac{x+\frac{11}{4}}{3}}$ = y - $\frac{1}{2}$

→ Add  $\frac{1}{2}$ to both sides

∴ ± $\sqrt{\frac{x+\frac{11}{4}}{3}}$ + $\frac{1}{2}$ = y

→ Replace y by $f^{-1}$(x)

∴ $f^{-1}$(x) =  ± $\sqrt{\frac{x+\frac{11}{4}}{3}}$ + $\frac{1}{2}$

∴ The inverse of f(x) is $f^{-1}$(x) =  ± $\sqrt{\frac{x+\frac{11}{4}}{3}}$ + $\frac{1}{2}$