Find the sum 112.1+ 1.2

F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

34kurt

Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

$g(f(x)) = 3x^4 + 18x^2 + 18$

Checking for inverse functions

$f(x) = 3x^2 + 9$

Represent f(x) with y

$y = 3x^2 + 9$

Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

Note that the resulting value of g(x) is not the same as $g(x) = \dfrac{1}{3}x^2 - 9$

<em>Hence, f(x) and g(x) and not inverse functions</em>

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