1. **Evaluate** $f(-9)$ for the piecewise function: The piecewise function $f(x)$ is defined as: - $2x - 1$ if $-5 \leq x > -4$ - $x + 5$ if $-1 \leq x \leq 1$ - $x^2$ if $x > 4$ Since $-9$ does not satisfy any of these intervals, $f(-9)$ is **not defined** in the given piecewise function. 2. **Calculate** $(f - g)(-1)$ where: - $f(x) = 2x^2$ - $g(x) = x^2 + 3x$ Step 1: Find $f(-1)$: $$f(-1) = 2(-1)^2 = 2 \times 1 = 2$$ Step 2: Find $g(-1)$: $$g(-1) = (-1)^2 + 3(-1) = 1 - 3 = -2$$ Step 3: Compute $(f - g)(-1)$: $$ (f - g)(-1) = f(-1) - g(-1) = 2 - (-2) = 2 + 2 = 4 $$ 3. **Find** $(f \circ g)(x)$ (composition of $f$ and $g$): By definition: $$ (f \circ g)(x) = f(g(x)) $$ Step 1: Write $g(x)$: $$ g(x) = x^2 + 3x $$ Step 2: Substitute $g(x)$ into $f$: $$ f(g(x)) = 2(g(x))^2 = 2(x^2 + 3x)^2 $$ Step 3: Expand $(x^2 + 3x)^2$: $$ (x^2 + 3x)^2 = (x^2)^2 + 2 \times x^2 \times 3x + (3x)^2 = x^4 + 6x^3 + 9x^2 $$ Step 4: Multiply by 2: $$ 2(x^4 + 6x^3 + 9x^2) = 2x^4 + 12x^3 + 18x^2 $$ **Final answers:** - $f(-9)$ is not defined in the piecewise function. - $(f - g)(-1) = 4$ - $(f \circ g)(x) = 2x^4 + 12x^3 + 18x^2$