1. **Problem:** Given $f(x) = x^2 + 3x - 7$, find $f(-2)$, $f(0)$, $f(4)$, $f(3x)$, and $f(2y)$. Step 1: Calculate $f(-2)$. Substitute $-2$ into $f(x)$: $$f(-2) = (-2)^2 + 3(-2) - 7 = 4 - 6 - 7 = -9$$ Step 2: Calculate $f(0)$: $$f(0) = 0^2 + 3(0) - 7 = 0 + 0 - 7 = -7$$ Step 3: Calculate $f(4)$: $$f(4) = 4^2 + 3(4) - 7 = 16 + 12 - 7 = 21$$ Step 4: Find $f(3x)$ by substituting $3x$ into $f(x)$: $$f(3x) = (3x)^2 + 3(3x) - 7 = 9x^2 + 9x - 7$$ Step 5: Find $f(2y)$: $$f(2y) = (2y)^2 + 3(2y) - 7 = 4y^2 + 6y - 7$$ 2. **Problem:** Given $g(x) = \sin 2x - \cos x$, find $g(\pi)$, $g(\frac{\pi}{2})$, $g(\pi + x)$, $g(-x)$, and $g(0)$. Step 1: Calculate $g(\pi)$: $$g(\pi) = \sin (2\pi) - \cos (\pi) = 0 - (-1) = 1$$ Step 2: Calculate $g(\frac{\pi}{2})$: $$g\left(\frac{\pi}{2}\right) = \sin \pi - \cos \frac{\pi}{2} = 0 - 0 = 0$$ Step 3: Calculate $g(\pi + x)$ using angle addition: $$g(\pi + x) = \sin 2(\pi + x) - \cos(\pi + x) = \sin(2\pi + 2x) - \cos \pi \cos x + \sin \pi \sin x$$ Since $\sin(2\pi + 2x) = \sin 2x$ and $\cos(\pi + x) = -\cos x$, $$g(\pi + x) = \sin 2x - (-\cos x) = \sin 2x + \cos x$$ Step 4: Calculate $g(-x)$: $$g(-x) = \sin(-2x) - \cos(-x) = -\sin 2x - \cos x$$ Step 5: Calculate $g(0)$: $$g(0) = \sin 0 - \cos 0 = 0 - 1 = -1$$ 3. **Problem:** Given $f(x) = 2x^2 - x + 5$ and $g(x) = 5x - 1$, find $(f \circ g)(2)$, $(f \circ y)(x)$, $(g \circ f)(x)$, and $(g \circ g)(x)$. Step 1: Find $(f \circ g)(2) = f(g(2))$: First, calculate $g(2)$: $$g(2) = 5(2) - 1 = 10 - 1 = 9$$ Then, $$f(9) = 2(9)^2 - 9 + 5 = 2(81) - 9 + 5 = 162 - 9 + 5 = 158$$ Step 2: Assume $(f \circ y)(x)$ means $(f \circ g)(x) = f(g(x))$; find $f(g(x))$: $$g(x) = 5x -1$$ $$f(g(x)) = 2(5x - 1)^2 - (5x -1) + 5$$ Expand: $$(5x - 1)^2 = 25x^2 - 10x + 1$$ So, $$f(g(x)) = 2(25x^2 - 10x + 1) - 5x + 1 + 5 = 50x^2 - 20x + 2 - 5x + 1 + 5$$ Simplify: $$50x^2 - 25x + 8$$ Step 3: Find $(g \circ f)(x) = g(f(x))$: $$f(x) = 2x^2 - x + 5$$ $$g(f(x)) = 5(2x^2 - x + 5) - 1 = 10x^2 - 5x + 25 - 1 = 10x^2 - 5x + 24$$ Step 4: Find $(g \circ g)(x) = g(g(x))$: $$g(x) = 5x - 1$$ $$g(g(x)) = g(5x - 1) = 5(5x - 1) - 1 = 25x - 5 - 1 = 25x - 6$$ **Final answers:** 1. $f(-2) = -9$, $f(0) = -7$, $f(4) = 21$, $f(3x) = 9x^2 + 9x - 7$, $f(2y) = 4y^2 + 6y - 7$ 2. $g(\pi) = 1$, $g(\frac{\pi}{2}) = 0$, $g(\pi + x) = \sin 2x + \cos x$, $g(-x) = -\sin 2x - \cos x$, $g(0) = -1$ 3. $(f \circ g)(2) = 158$, $(f \circ g)(x) = 50x^2 - 25x + 8$, $(g \circ f)(x) = 10x^2 - 5x + 24$, $(g \circ g)(x) = 25x - 6$