Answer :
Answer:
The inverse of f(x) is [tex]f^{-1}[/tex](x) = ± [tex]\sqrt{\frac{x+\frac{11}{4}}{3}}[/tex] + [tex]\frac{1}{2}[/tex]
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 = [tex]\frac{-b}{2a}[/tex]
- 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 = [tex]\frac{-(-3)}{2(3)}[/tex] = [tex]\frac{3}{6}[/tex] = [tex]\frac{1}{2}[/tex]
∴ h = [tex]\frac{1}{2}[/tex]
→ Substitute x by the value of h in f to find k
∵ k = 3( [tex]\frac{1}{2}[/tex])² - 3( [tex]\frac{1}{2}[/tex]) - 2
∴ k = [tex]-\frac{11}{4}[/tex]
→ Substitute the values of a, h, and k in the vertex form above
∵ f(x) = 3(x - [tex]\frac{1}{2}[/tex])² + [tex]-\frac{11}{4}[/tex]
∴ f(x) = 3(x - [tex]\frac{1}{2}[/tex])² - [tex]\frac{11}{4}[/tex]
Now let us find the inverse of f(x)
∵ f(x) = y
∴ y = 3(x - [tex]\frac{1}{2}[/tex])² - [tex]\frac{11}{4}[/tex]
→ Switch x and y
∵ x = 3(y - [tex]\frac{1}{2}[/tex])² - [tex]\frac{11}{4}[/tex]
→ Add [tex]\frac{11}{4}[/tex] to both sides
∴ x + [tex]\frac{11}{4}[/tex] = 3(y - [tex]\frac{1}{2}[/tex])²
→ Divide both sides by 3
∵ [tex]\frac{x+\frac{11}{4}}{3}[/tex] = (y - [tex]\frac{1}{2}[/tex])²
→ Take √ for both sides
∴ ± [tex]\sqrt{\frac{x+\frac{11}{4}}{3}}[/tex] = y - [tex]\frac{1}{2}[/tex]
→ Add [tex]\frac{1}{2}[/tex] to both sides
∴ ± [tex]\sqrt{\frac{x+\frac{11}{4}}{3}}[/tex] + [tex]\frac{1}{2}[/tex] = y
→ Replace y by [tex]f^{-1}[/tex](x)
∴ [tex]f^{-1}[/tex](x) = ± [tex]\sqrt{\frac{x+\frac{11}{4}}{3}}[/tex] + [tex]\frac{1}{2}[/tex]
∴ The inverse of f(x) is [tex]f^{-1}[/tex](x) = ± [tex]\sqrt{\frac{x+\frac{11}{4}}{3}}[/tex] + [tex]\frac{1}{2}[/tex]