1. **Problem Statement:** Find the extreme values (absolute and local) of the function $y=2x^2 - 8x + 9$ over its natural domain. 2. **Formula and Rules:** To find extreme values, we first find the critical points by setting the derivative equal to zero: $$y' = \frac{dy}{dx}$$. 3. **Calculate the derivative:** $$y' = \frac{d}{dx}(2x^2 - 8x + 9) = 4x - 8$$. 4. **Find critical points:** Set derivative equal to zero: $$4x - 8 = 0 \implies 4x = 8 \implies x = 2$$. 5. **Determine the nature of critical points:** Use the second derivative test: $$y'' = \frac{d}{dx}(4x - 8) = 4$$. Since $y'' = 4 > 0$, the function is concave up at $x=2$, so this point is a local minimum. 6. **Find the value of $y$ at $x=2$:** $$y(2) = 2(2)^2 - 8(2) + 9 = 8 - 16 + 9 = 1$$. 7. **Check endpoints and domain:** The domain is all real numbers, so no endpoints. 8. **Conclusion:** The function has a local (and absolute) minimum at $x=2$ with value $y=1$. There is no maximum since the parabola opens upward. **Final answer:** Local and absolute minimum at $(2,1)$. --- Repeat similar steps for each function: **54.** $y = x^3 - 2x + 4$ - Derivative: $y' = 3x^2 - 2$ - Critical points: $3x^2 - 2 = 0 \implies x = \pm \sqrt{\frac{2}{3}}$ - Second derivative: $y'' = 6x$ - At $x=\sqrt{2/3}$, $y'' > 0$ local min; at $x=-\sqrt{2/3}$, $y'' < 0$ local max. **55.** $y = x^3 + x^2 - 8x + 5$ - Derivative: $y' = 3x^2 + 2x - 8$ - Solve $3x^2 + 2x - 8=0$ for critical points. - Second derivative: $y'' = 6x + 2$ **56.** $y = x^3(x-5)^2$ - Expand or use product rule to find $y'$. - Find critical points by solving $y' = 0$. **57.** $y = \sqrt{x^2 - 1}$ - Domain: $|x| \geq 1$ - Derivative: $y' = \frac{x}{\sqrt{x^2 - 1}}$ - Critical points where numerator zero or derivative undefined. **58.** $y = x - 4\sqrt{x}$ - Domain: $x \geq 0$ - Derivative: $y' = 1 - \frac{2}{\sqrt{x}}$ - Solve $y' = 0$ for critical points. **59.** $y = \frac{1}{\sqrt{1 - x^2}}$ - Domain: $|x| < 1$ - Derivative: Use chain and quotient rules. **60.** $y = \sqrt{3} + 2x - x^2$ - Derivative: $y' = 2 - 2x$ - Critical point at $x=1$. **61.** $y = \frac{x}{x^2 + 1}$ - Derivative: Use quotient rule. **62.** $y = \frac{x+1}{x^2 + 2x + 2}$ - Derivative: Use quotient rule. **63.** $y = e^x + e^{-x}$ - Derivative: $y' = e^x - e^{-x}$ - Critical point at $x=0$. **64.** $y = e^x - e^{-x}$ - Derivative: $y' = e^x + e^{-x} > 0$ always, no critical points. **65.** $y = x \ln x$ - Domain: $x > 0$ - Derivative: $y' = \ln x + 1$ - Critical point at $x = e^{-1}$. **66.** $y = x^2 \ln x$ - Domain: $x > 0$ - Derivative: $y' = 2x \ln x + x$ - Solve $y' = 0$. **67.** $y = \cos^{-1}(x^2)$ - Domain: $|x| \leq 1$ - Derivative: $y' = \frac{-2x}{\sqrt{1 - x^4}}$ **68.** $y = \sin^{-1}(e^x)$ - Domain: $e^x \in [-1,1] \implies x \leq 0$ - Derivative: $y' = \frac{e^x}{\sqrt{1 - e^{2x}}}$ **Summary:** Each function's critical points are found by setting $y' = 0$ and analyzing $y''$ or using first derivative test to classify extrema. Absolute extrema depend on domain boundaries.