1. **Problem 1: Find the instantaneous rate of change of** $f(x) = 3x^2 - 6x$ **at** $x=1$. The instantaneous rate of change of a function at a point is given by the derivative evaluated at that point. 2. **Formula:** $$f'(x) = \frac{d}{dx}(3x^2 - 6x)$$ 3. **Calculate the derivative:** $$f'(x) = 6x - 6$$ 4. **Evaluate at** $x=1$: $$f'(1) = 6(1) - 6 = 6 - 6 = 0$$ So, the instantaneous rate of change at $x=1$ is $0$. --- 5. **Problem 2: Analyze the function** $f(x) = - \frac{x}{x^2 + 4}$ **and determine which statement is not true:** - $f$ has a local maximum at $x = -2$. - $f$ has a local minimum at $x = 2$. - $f$ is decreasing on $(-2, 2)$. - $f$ is increasing on $(-2, \infty)$. 6. **Find the derivative to analyze increasing/decreasing and extrema:** $$f(x) = - \frac{x}{x^2 + 4}$$ Use quotient rule: $$f'(x) = - \frac{(1)(x^2 + 4) - x(2x)}{(x^2 + 4)^2} = - \frac{x^2 + 4 - 2x^2}{(x^2 + 4)^2} = - \frac{-x^2 + 4}{(x^2 + 4)^2} = \frac{x^2 - 4}{(x^2 + 4)^2}$$ 7. **Critical points where** $f'(x) = 0$: $$x^2 - 4 = 0 \implies x = \pm 2$$ 8. **Sign of** $f'(x)$: - For $x$ in $(-2, 2)$, $x^2 - 4 < 0$ so $f'(x) < 0$ (decreasing). - For $x > 2$ or $x < -2$, $f'(x) > 0$ (increasing). 9. **Check statements:** - Local max at $x = -2$: Since $f'(x)$ changes from positive to negative at $x=-2$, this is true. - Local min at $x = 2$: Since $f'(x)$ changes from negative to positive at $x=2$, this is true. - $f$ is decreasing on $(-2, 2)$: True. - $f$ is increasing on $(-2, \infty)$: False, because $f'(x) < 0$ on $(-2, 2)$ and only positive on $(2, \infty)$. So, the false statement is: **$f$ is increasing on $(-2, \infty)$**. --- 10. **Problem 3: Given relation** $R = \{(x,y): y = 2x - 3\}$ **with domain** $\{-2, -1, 1, 3\}$, find the domain of** $R^{-1}$. 11. **Calculate the range of** $R$ for given domain: $$y = 2x - 3$$ For $x = -2$: $y = 2(-2) - 3 = -4 - 3 = -7$ For $x = -1$: $y = 2(-1) - 3 = -2 - 3 = -5$ For $x = 1$: $y = 2(1) - 3 = 2 - 3 = -1$ For $x = 3$: $y = 2(3) - 3 = 6 - 3 = 3$ 12. **The inverse relation** $R^{-1}$ **swaps** $x$ **and** $y$, so the domain of $R^{-1}$ is the range of $R$: $$\{-7, -5, -1, 3\}$$ --- **Final answers:** - Instantaneous rate of change at $x=1$ is $0$. - The false statement about $f(x) = - \frac{x}{x^2 + 4}$ is: "$f$ is increasing on $(-2, \infty)$". - The domain of $R^{-1}$ is $\{-7, -5, -1, 3\}$.