1. Given the point (4, 10) is a maximum on the quadratic function $f(x)$, analyze the statements: 1. The average rate of change over $[0,4]$ is positive. - The average rate of change is given by $$\frac{f(4)-f(0)}{4-0}.$$ Since $(4,10)$ is a maximum, $f(4)$ is the highest value in that region, so $f(4) > f(0)$, making the average rate of change positive.\ 2. The instantaneous rate of change at $x=4$ is 0. - At a maximum point of a differentiable function, the derivative (instantaneous rate of change) is zero. So this is true. 3. $f(6) > f(4)$ - Since $(4,10)$ is a maximum, values of $f(x)$ near 4 but not equal to 4 are less than or equal to 10. So $f(6) < f(4)$, making this false. 4. There is only one possible $x$ with instantaneous rate of change 0. - A quadratic function has at most two critical points. Since $(4,10)$ is a maximum, the other critical point is a minimum. So there are two values where derivative is zero. This statement is false. 2. Prove $g(x) = x^3 - 2x^2 - 15x + 12$ has a local minimum at $(3,-24)$ using rates of change: - First, find $g'(x) = 3x^2 - 4x - 15$. - Evaluate $g'(3) = 3(9) - 4(3) - 15 = 27 - 12 - 15 = 0$, so slope is zero at $x=3$. - Find $g''(x) = 6x - 4$. - Evaluate $g''(3) = 18 - 4 = 14 > 0$, so the function is concave up at $x=3$, confirming a local minimum. 3. Find $a$ so average rate of change of $f(x) = 3x^2 + ax - 4$ over $[-1,2]$ is 8: - Average rate of change = $$\frac{f(2) - f(-1)}{2 - (-1)} = 8$$ - Calculate $f(2) = 3(4) + 2a - 4 = 12 + 2a - 4 = 8 + 2a$ - Calculate $f(-1) = 3(1) - a - 4 = 3 - a - 4 = -1 - a$ - Substitute: $$\frac{(8 + 2a) - (-1 - a)}{3} = 8$$ - Simplify numerator: $8 + 2a + 1 + a = 9 + 3a$ - Equation: $$\frac{9 + 3a}{3} = 8$$ - Multiply both sides by 3: $$9 + 3a = 24$$ - Subtract 9: $$3a = 15$$ - Divide by 3: $$a = 5$$ 4. For $f(x) = \frac{4}{x-3}$, find $x$ where tangent slope equals secant slope $-\frac{1}{16}$: - Derivative: $$f'(x) = -\frac{4}{(x-3)^2}$$ - Set $f'(x) = -\frac{1}{16}$: $$-\frac{4}{(x-3)^2} = -\frac{1}{16}$$ - Multiply both sides by $(x-3)^2$: $$-4 = -\frac{(x-3)^2}{16}$$ - Multiply both sides by $-16$: $$64 = (x-3)^2$$ - Take square root: $$x-3 = \pm 8$$ - So $x = 3 + 8 = 11$ or $x = 3 - 8 = -5$ 5. Sinusoidal function graph analysis: a. Model equation: - Amplitude $A = \frac{20 - 10}{2} = 5$ (approximate from peak and trough) - Midline $D = \frac{20 + 10}{2} = 15$ - Period $T \approx 20$ (distance between peaks near $x=5$ and $x=25$) - Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{20} = \frac{\pi}{10}$ - Phase shift $\phi = 0$ (starts near max at $x=0$) - Equation: $$f(x) = 5 \cos\left(\frac{\pi}{10} x\right) + 15$$ b. Interval where rate of change is negative: - Rate of change is derivative: $$f'(x) = -5 \cdot \frac{\pi}{10} \sin\left(\frac{\pi}{10} x\right) = -\frac{\pi}{2} \sin\left(\frac{\pi}{10} x\right)$$ - Derivative negative when $\sin\left(\frac{\pi}{10} x\right) > 0$. - This occurs on intervals where $\frac{\pi}{10} x \in (0, \pi)$, so $$x \in (0, 10)$$ c. Point where instantaneous rate of change is maximum: - Maximum rate of change magnitude occurs when $|\sin(\frac{\pi}{10} x)| = 1$. - Maximum positive rate of change when $\sin(\frac{\pi}{10} x) = -1$ (because of negative sign in derivative). - Solve $\sin(\frac{\pi}{10} x) = -1$: $$\frac{\pi}{10} x = \frac{3\pi}{2} \Rightarrow x = 15$$ - Rate of change at $x=15$: $$f'(15) = -\frac{\pi}{2} \sin\left(\frac{\pi}{10} \times 15\right) = -\frac{\pi}{2} \times (-1) = \frac{\pi}{2} \approx 1.5708$$ - Point on graph: $$f(15) = 5 \cos\left(\frac{\pi}{10} \times 15\right) + 15 = 5 \cos\left(\frac{3\pi}{2}\right) + 15 = 5 \times 0 + 15 = 15$$ Final answers: - a: $f(x) = 5 \cos\left(\frac{\pi}{10} x\right) + 15$ - b: Rate of change negative on $(0,10)$ - c: Instantaneous rate of change max at $(15,15)$ with value $\frac{\pi}{2}$