1. **State the problem:** We have the function $f(x) = 2x^2 - 8x + 1$ defined on the interval $0 \leq x \leq 4$. We need to find: - Critical points - Absolute maximum and minimum values on the interval - Intervals where the function is increasing or decreasing 2. **Find critical points:** Critical points occur where the derivative $f'(x)$ is zero or undefined. Calculate the derivative: $$f'(x) = \frac{d}{dx}(2x^2 - 8x + 1) = 4x - 8$$ Set derivative equal to zero: $$4x - 8 = 0 \implies 4x = 8 \implies x = 2$$ Since $x=2$ is within the interval $[0,4]$, it is a critical point. 3. **Find absolute max and min on $[0,4]$:** Evaluate $f(x)$ at critical points and endpoints: - At $x=0$: $f(0) = 2(0)^2 - 8(0) + 1 = 1$ - At $x=2$: $f(2) = 2(2)^2 - 8(2) + 1 = 2(4) - 16 + 1 = 8 - 16 + 1 = -7$ - At $x=4$: $f(4) = 2(4)^2 - 8(4) + 1 = 2(16) - 32 + 1 = 32 - 32 + 1 = 1$ The absolute minimum is $-7$ at $x=2$. The absolute maximum is $1$ at both $x=0$ and $x=4$. 4. **Determine intervals of increase and decrease:** Recall $f'(x) = 4x - 8$. - For $x < 2$, say $x=1$, $f'(1) = 4(1) - 8 = -4 < 0$, so $f$ is decreasing on $(0,2)$. - For $x > 2$, say $x=3$, $f'(3) = 4(3) - 8 = 12 - 8 = 4 > 0$, so $f$ is increasing on $(2,4)$. 5. **Find the slope of $y^2 + x^2 = 1$ at $(1,1)$:** Implicitly differentiate both sides with respect to $x$: $$2y \frac{dy}{dx} + 2x = 0$$ Solve for $\frac{dy}{dx}$: $$2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$$ At point $(1,1)$: $$\frac{dy}{dx} = -\frac{1}{1} = -1$$ **Final answers:** - Critical point: $x=2$ - Absolute minimum: $f(2) = -7$ - Absolute maximum: $f(0) = f(4) = 1$ - Increasing on $(2,4)$ - Decreasing on $(0,2)$ - Slope of $y^2 + x^2 = 1$ at $(1,1)$ is $-1$