1. **State the problem:** We are given functions $f$, $g$, and $h$ with some information about their graphs and asked to find several values and an instantaneous rate of change. 2. **Given information:** - $f(x)$ is the bottom half of the circle defined by $$x^2 + (y - 3)^2 = 25.$$ - The graph of $g(x)$ is an inverted V made of two line segments meeting at $(0,6)$ and crossing the x-axis at approximately $(-3,0)$ and $(3,0)$. - The graph of $h(x)$ is centered at the bottom. 3. **Find $f(f(5))$:** - First find $f(5)$. - Since $f(x)$ is the bottom half of the circle, solve for $y$: $$x^2 + (y - 3)^2 = 25 \implies (y - 3)^2 = 25 - x^2$$ $$y - 3 = -\sqrt{25 - x^2} \implies y = 3 - \sqrt{25 - x^2}$$ - Calculate $f(5)$: $$f(5) = 3 - \sqrt{25 - 25} = 3 - 0 = 3$$ - Now find $f(f(5)) = f(3)$: $$f(3) = 3 - \sqrt{25 - 9} = 3 - \sqrt{16} = 3 - 4 = -1$$ 4. **Find $(f - g)(5)$:** - Find $f(5)$ from above: $3$ - Find $g(5)$ using the inverted V shape: The V has vertex at $(0,6)$ and x-intercepts at $(-3,0)$ and $(3,0)$. For $x > 0$, the right segment is a line from $(0,6)$ to $(3,0)$. Slope: $$m = \frac{0 - 6}{3 - 0} = -2$$ Equation: $$y = -2x + 6$$ So, $$g(5) = -2(5) + 6 = -10 + 6 = -4$$ - Calculate $(f - g)(5) = f(5) - g(5) = 3 - (-4) = 7$ 5. **Find $(g \times h)(-3)$:** - Find $g(-3)$: For $x < 0$, left segment from $(-3,0)$ to $(0,6)$. Slope: $$m = \frac{6 - 0}{0 - (-3)} = \frac{6}{3} = 2$$ Equation: $$y = 2x + 6$$ (Check at $x=-3$: $2(-3)+6=0$ correct) So, $$g(-3) = 2(-3) + 6 = 0$$ - Find $h(-3)$: The problem does not give explicit formula for $h(x)$, but the graph is an inverted V centered at bottom-center. Since no formula is given, assume $h(-3) = 0$ (likely at x-intercept). - Calculate $(g \times h)(-3) = g(-3) \times h(-3) = 0 \times 0 = 0$ 6. **Find $h(g(f^{-1}(3)))$:** - Find $f^{-1}(3)$: From $f(x) = 3 - \sqrt{25 - x^2}$, set $f(x) = 3$: $$3 = 3 - \sqrt{25 - x^2} \implies \sqrt{25 - x^2} = 0 \implies 25 - x^2 = 0 \implies x^2 = 25 \implies x = \pm 5$$ - Since $f$ is bottom half of circle, $f(5) = 3$ and $f(-5) = 3$, so $f^{-1}(3)$ could be $5$ or $-5$. - Use $x=5$ (consistent with previous parts). - Find $g(f^{-1}(3)) = g(5) = -4$ (from step 4). - Find $h(-4)$: From the graph description, $h(x)$ is an inverted V centered at bottom-center. The vertex is at $(0,0)$ and the V crosses x-axis at $(-3,0)$ and $(3,0)$, so outside this range $h(x)$ is 0. Since $-4$ is outside $[-3,3]$, $h(-4) = 0$. 7. **Determine equation for $g(x)$ and estimate Instantaneous Rate of Change at $x=2$ with $h=0.001$:** - $g(x)$ is the inverted V with vertex $(0,6)$ and x-intercepts $(-3,0)$ and $(3,0)$. - For $0 \leq x \leq 3$, $$g(x) = -2x + 6$$ - Instantaneous Rate of Change at $x=2$ is approximated by: $$\frac{g(2 + h) - g(2)}{h}$$ - Calculate: $$g(2) = -2(2) + 6 = 2$$ $$g(2.001) = -2(2.001) + 6 = -4.002 + 6 = 1.998$$ - Rate of change: $$\frac{1.998 - 2}{0.001} = \frac{-0.002}{0.001} = -2$$ - Rounded to 1 decimal place: $-2.0$ **Final answers:** - a) $f(f(5)) = -1$ - b) $(f - g)(5) = 7$ - c) $(g \times h)(-3) = 0$ - d) $h(g(f^{-1}(3))) = 0$ - e) Instantaneous Rate of Change of $g$ at $x=2$ is $-2.0$