1. **Problem:** Find the value(s) of $c$ that satisfy the Mean Value Theorem (MVT) for $f(x) = x^2 + 2x - 1$ on $[0,1]$. **Step 1:** State the MVT formula: $$\frac{f(b) - f(a)}{b - a} = f'(c)$$ where $a=0$, $b=1$. **Step 2:** Calculate $f(a)$ and $f(b)$: $$f(0) = 0^2 + 2\cdot0 - 1 = -1$$ $$f(1) = 1^2 + 2\cdot1 - 1 = 1 + 2 - 1 = 2$$ **Step 3:** Compute the average rate of change: $$\frac{f(1) - f(0)}{1 - 0} = \frac{2 - (-1)}{1} = 3$$ **Step 4:** Find $f'(x)$: $$f'(x) = 2x + 2$$ **Step 5:** Set $f'(c) = 3$ and solve for $c$: $$2c + 2 = 3 \implies 2c = 1 \implies c = \frac{1}{2}$$ --- 2. **Problem:** Find $c$ for $f(x) = x^{2/3}$ on $[0,1]$. **Step 1:** Calculate $f(0)$ and $f(1)$: $$f(0) = 0^{2/3} = 0$$ $$f(1) = 1^{2/3} = 1$$ **Step 2:** Average rate of change: $$\frac{1 - 0}{1 - 0} = 1$$ **Step 3:** Derivative: $$f'(x) = \frac{2}{3} x^{-1/3}$$ Note: $f'(0)$ is undefined. **Step 4:** Solve $f'(c) = 1$: $$\frac{2}{3} c^{-1/3} = 1 \implies c^{-1/3} = \frac{3}{2} \implies c^{1/3} = \frac{2}{3} \implies c = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$$ --- 3. **Problem:** Find $c$ for $f(x) = x + \frac{1}{x}$ on $[\frac{1}{2}, 2]$. **Step 1:** Calculate $f(\frac{1}{2})$ and $f(2)$: $$f\left(\frac{1}{2}\right) = \frac{1}{2} + 2 = \frac{5}{2}$$ $$f(2) = 2 + \frac{1}{2} = \frac{5}{2}$$ **Step 2:** Average rate of change: $$\frac{\frac{5}{2} - \frac{5}{2}}{2 - \frac{1}{2}} = 0$$ **Step 3:** Derivative: $$f'(x) = 1 - \frac{1}{x^2}$$ **Step 4:** Solve $f'(c) = 0$: $$1 - \frac{1}{c^2} = 0 \implies \frac{1}{c^2} = 1 \implies c^2 = 1 \implies c = \pm 1$$ Since $c \in \left(\frac{1}{2}, 2\right)$, $c=1$. --- 4. **Problem:** Find $c$ for $f(x) = \sqrt{x - 1}$ on $[1,3]$. **Step 1:** Calculate $f(1)$ and $f(3)$: $$f(1) = \sqrt{0} = 0$$ $$f(3) = \sqrt{2}$$ **Step 2:** Average rate of change: $$\frac{\sqrt{2} - 0}{3 - 1} = \frac{\sqrt{2}}{2}$$ **Step 3:** Derivative: $$f'(x) = \frac{1}{2\sqrt{x - 1}}$$ **Step 4:** Solve $f'(c) = \frac{\sqrt{2}}{2}$: $$\frac{1}{2\sqrt{c - 1}} = \frac{\sqrt{2}}{2} \implies \frac{1}{\sqrt{c - 1}} = \sqrt{2} \implies \sqrt{c - 1} = \frac{1}{\sqrt{2}}$$ Square both sides: $$c - 1 = \frac{1}{2} \implies c = \frac{3}{2}$$ --- 5. **Problem:** Check if $f(x) = x^{2/3}$ satisfies MVT on $[-1,8]$. **Step 1:** Check continuity on $[-1,8]$: $f(x)$ is continuous for all real $x$. **Step 2:** Check differentiability on $(-1,8)$: $$f'(x) = \frac{2}{3} x^{-1/3}$$ Not differentiable at $x=0$ (since derivative undefined). **Conclusion:** Does not satisfy MVT hypotheses because $f$ is not differentiable on entire open interval. --- 6. **Problem:** Check $f(x) = x^{4/5}$ on $[0,1]$. **Step 1:** Continuity: $f$ continuous on $[0,1]$. **Step 2:** Differentiability: $$f'(x) = \frac{4}{5} x^{-1/5}$$ Derivative undefined at $x=0$. **Conclusion:** Does not satisfy MVT hypotheses. --- 7. **Problem:** Check $f(x) = \sqrt{x(1-x)}$ on $[0,1]$. **Step 1:** Continuity: $f$ continuous on $[0,1]$. **Step 2:** Differentiability: $$f'(x) = \frac{1 - 2x}{2\sqrt{x(1-x)}}$$ Derivative undefined at $x=0$ and $x=1$ (division by zero). **Conclusion:** Differentiable on $(0,1)$ but not at endpoints. MVT requires differentiability on open interval, so it satisfies hypotheses. --- 8. **Problem:** Check $f(x) = \begin{cases} \frac{\sin x}{x}, & -\pi \leq x < 0 \\ 0, & x=0 \end{cases}$ on $[-\pi,0]$. **Step 1:** Continuity at $x=0$: $$\lim_{x \to 0^-} \frac{\sin x}{x} = 1 \neq f(0) = 0$$ **Conclusion:** Not continuous at $x=0$, so MVT hypotheses fail. --- **Summary:** - Problems 1-4: Found $c$ values satisfying MVT. - Problems 5-8: Only 7 satisfies MVT hypotheses.