\( 1 \leftarrow \quad \) 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. \( (f \circ g)(x)=-2 x^{2}-x-1 \) (Simplify your answer.) b. \( (g \circ f)(x)=\square \) (Simplify your answer.)

To find \( (g \circ f)(x) \), we need to substitute \( f(x) \) into \( g(x) \). So, we start with: \[ g(f(x)) = g(1 - x) \] Now, substitute \( f(x) = 1 - x \) into \( g(x) = 2x^2 + x + 2 \): \[ g(1 - x) = 2(1 - x)^2 + (1 - x) + 2 \] \[ = 2(1 - 2x + x^2) + (1 - x) + 2 \] \[ = 2 - 4x + 2x^2 + 1 - x + 2 \] \[ = 2x^2 - 5x + 5 \] Thus, b. \( (g \circ f)(x) = 2x^2 - 5x + 5 \). Now, let's evaluate \( (f \circ g)(3) \) and \( (g \circ f)(3) \): For \( (f \circ g)(3) \), we first find \( g(3) \): \[ g(3) = 2(3)^2 + (3) + 2 = 2(9) + 3 + 2 = 18 + 3 + 2 = 23 \] Now, substitute this into \( f \): \[ f(23) = 1 - 23 = -22 \] Therefore, c. \( (f \circ g)(3) = -22 \). Next, let’s find \( (g \circ f)(3) \): We know \( f(3) = 1 - 3 = -2 \). Now we substitute \(-2\) into \( g \): \[ g(-2) = 2(-2)^2 + (-2) + 2 = 2(4) - 2 + 2 = 8 - 2 + 2 = 8 \] Thus, d. \( (g \circ f)(3) = 8 \). In summary: - \( (f \circ g)(x) = -2x^2 - x - 1 \) - \( (g \circ f)(x) = 2x^2 - 5x + 5 \) - \( (f \circ g)(3) = -22 \) - \( (g \circ f)(3) = 8 \)