Problem

ons of Functions
For $f(x)=1-x$ and $g(x)=4 x^{2}+x+4$, find the following functions
a. $(f \circ g)(x) ; b .(g \circ f)(x) ; c .(f \circ g)(2) ; d .(g \circ f)(2)$

Answer

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Answer

Final Answer: \(\boxed{(f \circ g)(x) = -4x^2 - x - 3}\), \(\boxed{(g \circ f)(x) = 4x^2 - 9x + 9}\), \(\boxed{(f \circ g)(2) = -21}\), \(\boxed{(g \circ f)(2) = 7}\)

Steps

Step 1 :Given functions are \(f(x) = 1 - x\) and \(g(x) = 4x^2 + x + 4\)

Step 2 :The composition of two functions, denoted as \((f \circ g)(x)\), is the function obtained by applying \(f\) to the result of applying \(g\) to \(x\). In other words, \((f \circ g)(x) = f(g(x))\). Similarly, \((g \circ f)(x) = g(f(x))\).

Step 3 :To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). So, \((f \circ g)(x) = f(g(x)) = f(4x^2 + x + 4) = 1 - (4x^2 + x + 4) = -4x^2 - x + 1 - 4 = -4x^2 - x - 3\)

Step 4 :Similarly, to find \((g \circ f)(x)\), we need to substitute \(f(x)\) into \(g(x)\). So, \((g \circ f)(x) = g(f(x)) = g(1 - x) = 4(1 - x)^2 + (1 - x) + 4 = 4(1 - 2x + x^2) + 1 - x + 4 = 4x^2 - 8x + 4 + 1 - x + 4 = 4x^2 - 9x + 9\)

Step 5 :For part c, we need to substitute \(x = 2\) into the function \((f \circ g)(x)\). So, \((f \circ g)(2) = -4*2^2 - 2 - 3 = -4*4 - 2 - 3 = -16 - 2 - 3 = -21\)

Step 6 :For part d, we need to substitute \(x = 2\) into the function \((g \circ f)(x)\). So, \((g \circ f)(2) = 4*2^2 - 9*2 + 9 = 4*4 - 18 + 9 = 16 - 18 + 9 = 7\)

Step 7 :Final Answer: \(\boxed{(f \circ g)(x) = -4x^2 - x - 3}\), \(\boxed{(g \circ f)(x) = 4x^2 - 9x + 9}\), \(\boxed{(f \circ g)(2) = -21}\), \(\boxed{(g \circ f)(2) = 7}\)

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