Answer:
[tex]f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252[/tex]
[tex]g(f(x)) = 3x^4 + 18x^2 + 18[/tex]
f(x) and g(x) and not inverse functions
Step-by-step explanation:
Given
[tex]f(x) = 3x^2 + 9[/tex]
[tex]g(x) = \dfrac{1}{3}x^2 - 9[/tex]
Required
Determine f(g(x))
Determine g(f(x))
Determine if both functions are inverse:
Calculating f(g(x))
[tex]f(x) = 3x^2 + 9[/tex]
[tex]f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9[/tex]
[tex]f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9[/tex]
Expand Brackets
[tex]f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9[/tex]
[tex]f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9[/tex]
[tex]f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9[/tex]
[tex]f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252[/tex]
Calculating g(f(x))
[tex]g(x) = \dfrac{1}{3}x^2 - 9[/tex]
[tex]g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9[/tex]
[tex]g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9[/tex]
[tex]g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9[/tex]
Expand Brackets
[tex]g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9[/tex]
[tex]g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9[/tex]
[tex]g(f(x)) = 3x^4 + 18x^2 + 18[/tex]
Checking for inverse functions
[tex]f(x) = 3x^2 + 9[/tex]
Represent f(x) with y
[tex]y = 3x^2 + 9[/tex]
Swap positions of x and y
[tex]x = 3y^2 + 9[/tex]
Subtract 9 from both sides
[tex]x - 9 = 3y^2 + 9 - 9[/tex]
[tex]x - 9 = 3y^2[/tex]
[tex]3y^2 = x - 9[/tex]
Divide through by 3
[tex]\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}[/tex]
[tex]y^2 = \frac{x}{3} - 3[/tex]
Take square root of both sides
[tex]\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}[/tex]
[tex]y = \sqrt{\frac{x}{3} - 3}[/tex]
Represent y with g(x)
[tex]g(x) = \sqrt{\frac{x}{3} - 3}[/tex]
Note that the resulting value of g(x) is not the same as [tex]g(x) = \dfrac{1}{3}x^2 - 9[/tex]
Hence, f(x) and g(x) and not inverse functions
