\( 1 \leftarrow \) For \( f(x)=1-x \) and \( g(x)=2 x^{2}+x+2 \), find the following functions. a. \( (f \circ g)(x) ; b .(g \circ f)(x) ; \) c. \( (f \circ g)(3) ; \) d. \( (g \circ f)(3) \) a. \( (f \circ g)(x)=\square \) (Simplify your answer.)

To find \( (f \circ g)(x) \), we will substitute \( g(x) \) into \( f(x) \). Given: - \( f(x) = 1 - x \) - \( g(x) = 2x^{2} + x + 2 \) We have: \[ (f \circ g)(x) = f(g(x)) = f(2x^{2} + x + 2) \] Now, substitute \( g(x) \) into \( f(x) \): \[ f(2x^{2} + x + 2) = 1 - (2x^{2} + x + 2) \] This simplifies to: \[ = 1 - 2x^{2} - x - 2 \] \[ = -2x^{2} - x - 1 \] Thus, a. \( (f \circ g)(x) = -2x^{2} - x - 1 \) Now for part b, \( (g \circ f)(x) \): We need to substitute \( f(x) \) into \( g(x) \): \[ (g \circ f)(x) = g(f(x)) = g(1 - x) \] Now substitute \( f(x) \) into \( g(x) \): \[ g(1 - x) = 2(1 - x)^{2} + (1 - x) + 2 \] Calculating \( (1 - x)^{2} = 1 - 2x + x^{2} \): \[ = 2(1 - 2x + x^{2}) + 1 - x + 2 \] Distributing the 2: \[ = 2 - 4x + 2x^{2} + 1 - x + 2 \] Combining like terms: \[ = 2x^{2} - 5x + 5 \] Thus, b. \( (g \circ f)(x) = 2x^{2} - 5x + 5 \) Next, for part c, we calculate \( (f \circ g)(3) \): \[ (f \circ g)(3) = -2(3)^{2} - (3) - 1 \] \[ = -2(9) - 3 - 1 \] \[ = -18 - 3 - 1 \] \[ = -22 \] So, c. \( (f \circ g)(3) = -22 \) Finally, for part d, \( (g \circ f)(3) \): \[ (g \circ f)(3) = 2(3)^{2} - 5(3) + 5 \] \[ = 2(9) - 15 + 5 \] \[ = 18 - 15 + 5 \] \[ = 8 \] Therefore, d. \( (g \circ f)(3) = 8 \)