1. **State the problem:** We are given three functions: - $f(x)$ is the bottom half of the circle defined by $$x^2 + (y - 3)^2 = 25$$ - $g(x)$ is a sinusoidal function oscillating between 4 and -4 - $h(x)$ is an inverted V-shaped graph with vertex at $(0,6)$ and x-intercepts at $(-3,0)$ and $(3,0)$ We need to find: a) $f(f(5))$ b) $(f - g)(5)$ c) $(g \times h)(-3)$ d) $h(g(f^{-1}(3)))$ e) Equation for $g(x)$ and estimate the instantaneous rate of change at $x=2$ with $h=0.001$ --- 2. **Find $f(x)$ from the circle equation:** Given the circle equation: $$x^2 + (y - 3)^2 = 25$$ Solve for $y$: $$(y - 3)^2 = 25 - x^2$$ $$y - 3 = \pm \sqrt{25 - x^2}$$ Since $f(x)$ is the bottom half of the circle, take the negative root: $$f(x) = 3 - \sqrt{25 - x^2}$$ --- 3. **Calculate $f(5)$:** $$f(5) = 3 - \sqrt{25 - 5^2} = 3 - \sqrt{25 - 25} = 3 - 0 = 3$$ 4. **Calculate $f(f(5)) = f(3)$:** $$f(3) = 3 - \sqrt{25 - 3^2} = 3 - \sqrt{25 - 9} = 3 - \sqrt{16} = 3 - 4 = -1$$ So, $f(f(5)) = -1$ --- 5. **Calculate $(f - g)(5) = f(5) - g(5)$:** We know $f(5) = 3$. From the description, $g(x)$ oscillates between 4 and -4 with peaks near $x=5$. At $x=5$, $g(5)$ is near a peak, so $g(5) \approx 4$. Therefore: $$(f - g)(5) = 3 - 4 = -1$$ --- 6. **Calculate $(g \times h)(-3) = g(-3) \times h(-3)$:** - For $g(-3)$: The troughs are near $x=-5, -1, 3$. At $x=-3$, $g(x)$ is between trough at $-5$ and trough at $-1$, so near a peak or zero crossing. Since peaks are at $-2$ and $1$, $g(-3)$ is near zero crossing, estimate $g(-3) \approx 0$. - For $h(-3)$: The inverted V has vertex at $(0,6)$ and x-intercepts at $(-3,0)$ and $(3,0)$. At $x=-3$, $h(-3) = 0$. Therefore: $$(g \times h)(-3) = 0 \times 0 = 0$$ --- 7. **Calculate $h(g(f^{-1}(3)))$:** - First find $f^{-1}(3)$: Solve $f(x) = 3$. Recall $f(x) = 3 - \sqrt{25 - x^2}$. Set equal to 3: $$3 - \sqrt{25 - x^2} = 3$$ $$\sqrt{25 - x^2} = 0$$ $$25 - x^2 = 0$$ $$x^2 = 25$$ $$x = \pm 5$$ Since $f(x)$ is defined on $[-5,5]$, $f^{-1}(3)$ can be $5$ or $-5$. - Calculate $g(f^{-1}(3)) = g(5)$ or $g(-5)$. From earlier, $g(5) \approx 4$ (peak), and $g(-5)$ is a trough near $-4$. Choose $f^{-1}(3) = 5$ (positive root). So, $g(5) = 4$. - Calculate $h(4)$: The inverted V has x-intercepts at $-3$ and $3$, so domain is $[-3,3]$. At $x=4$, $h(4)$ is outside the domain, but the graph extends downward to about $(-6,-6)$ and $(6,-6)$. Assuming linear extension beyond $3$, $h(4)$ is approximately $-6$. Therefore: $$h(g(f^{-1}(3))) = h(4) \approx -6$$ --- 8. **Determine equation for $g(x)$:** Given $g(x)$ is sinusoidal with amplitude 4, peaks at $x=-2,1,5$. Period $T$ is distance between peaks: $1 - (-2) = 3$ or $5 - 1 = 4$ (approximate, take average $3.5$). Assuming period $T=3$, angular frequency: $$\omega = \frac{2\pi}{T} = \frac{2\pi}{3}$$ Phase shift $\phi$ to match peak at $x=-2$: $$\omega(-2) + \phi = \frac{\pi}{2}$$ $$\phi = \frac{\pi}{2} + 2\omega = \frac{\pi}{2} + 2 \times \frac{2\pi}{3} = \frac{\pi}{2} + \frac{4\pi}{3} = \frac{11\pi}{6}$$ Equation: $$g(x) = 4 \sin\left(\frac{2\pi}{3} x + \frac{11\pi}{6}\right)$$ --- 9. **Estimate instantaneous rate of change of $g(x)$ at $x=2$ using $h=0.001$:** Use difference quotient: $$\text{Rate} = \frac{g(2 + 0.001) - g(2)}{0.001}$$ Calculate: $$g(2) = 4 \sin\left(\frac{2\pi}{3} \times 2 + \frac{11\pi}{6}\right) = 4 \sin\left(\frac{4\pi}{3} + \frac{11\pi}{6}\right) = 4 \sin\left(\frac{8\pi}{6} + \frac{11\pi}{6}\right) = 4 \sin\left(\frac{19\pi}{6}\right)$$ Since $\sin(\theta)$ is periodic with period $2\pi$, $\frac{19\pi}{6} = 3\pi + \frac{\pi}{6}$, and $\sin(3\pi + \frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}$. So, $$g(2) = 4 \times \left(-\frac{1}{2}\right) = -2$$ Similarly, $$g(2.001) = 4 \sin\left(\frac{2\pi}{3} \times 2.001 + \frac{11\pi}{6}\right) = 4 \sin\left(\frac{4.002\pi}{3} + \frac{11\pi}{6}\right) = 4 \sin\left(\frac{8.004\pi}{6} + \frac{11\pi}{6}\right) = 4 \sin\left(\frac{19.004\pi}{6}\right)$$ Approximate: $$\sin\left(\frac{19.004\pi}{6}\right) \approx \sin\left(3\pi + \frac{\pi}{6} + 0.004\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6} + 0.004\frac{\pi}{6}\right)$$ Calculate inside sine: $$\frac{\pi}{6} + 0.004 \times \frac{\pi}{6} = \frac{\pi}{6} (1 + 0.004) = \frac{\pi}{6} \times 1.004$$ Use small angle approximation: $$\sin(a + \delta) \approx \sin a + \delta \cos a$$ So, $$\sin\left(\frac{\pi}{6} + 0.004 \times \frac{\pi}{6}\right) \approx \sin\left(\frac{\pi}{6}\right) + 0.004 \times \frac{\pi}{6} \cos\left(\frac{\pi}{6}\right) = \frac{1}{2} + 0.004 \times \frac{\pi}{6} \times \frac{\sqrt{3}}{2}$$ Calculate: $$0.004 \times \frac{\pi}{6} \times \frac{\sqrt{3}}{2} \approx 0.004 \times 0.5236 \times 0.866 = 0.0018$$ So, $$\sin\left(\frac{\pi}{6} + 0.004 \times \frac{\pi}{6}\right) \approx 0.5 + 0.0018 = 0.5018$$ Therefore, $$\sin\left(\frac{19.004\pi}{6}\right) \approx -0.5018$$ Calculate difference quotient: $$\frac{g(2.001) - g(2)}{0.001} = \frac{4 \times (-0.5018) - (-2)}{0.001} = \frac{-2.0072 + 2}{0.001} = \frac{-0.0072}{0.001} = -7.2$$ Rounded to 1 decimal place: $$-7.2$$ --- **Final answers:** - a) $f(f(5)) = -1$ - b) $(f - g)(5) = -1$ - c) $(g \times h)(-3) = 0$ - d) $h(g(f^{-1}(3))) \approx -6$ - e) Instantaneous rate of change of $g(x)$ at $x=2$ is approximately $-7.2$