1. Verify Rolle's theorem for $f(x) = x^2 - 3x + 4$ on $[1,2]$. Step 1: Check if $f(1) = f(2)$. Calculate $f(1) = 1^2 - 3(1) + 4 = 1 - 3 + 4 = 2$. Calculate $f(2) = 2^2 - 3(2) + 4 = 4 - 6 + 4 = 2$. Since $f(1) = f(2)$, the first condition of Rolle's theorem is satisfied. Step 2: Find $f'(x)$. $$f'(x) = 2x - 3$$ Step 3: Find $c$ in $(1,2)$ such that $f'(c) = 0$. Solve $2c - 3 = 0$. $$2c = 3 \Rightarrow c = \frac{3}{2} = 1.5$$ Since $1.5 \in (1,2)$, Rolle's theorem is verified. 2. Find all $c$ in $[0,2]$ satisfying the Mean Value Theorem for $f(x) = x^3 - 4x$. Step 1: Calculate $f(0)$ and $f(2)$. $f(0) = 0^3 - 4(0) = 0$ $f(2) = 2^3 - 4(2) = 8 - 8 = 0$ Step 2: Compute the average rate of change. $$\frac{f(2) - f(0)}{2 - 0} = \frac{0 - 0}{2} = 0$$ Step 3: Find $f'(x)$. $$f'(x) = 3x^2 - 4$$ Step 4: Solve $f'(c) = 0$. $$3c^2 - 4 = 0 \Rightarrow 3c^2 = 4 \Rightarrow c^2 = \frac{4}{3} \Rightarrow c = \pm \frac{2}{\sqrt{3}}$$ Only $c = \frac{2}{\sqrt{3}} \approx 1.1547$ lies in $[0,2]$. 3. Prove that between any two real roots of $e^x \sin x = 1$, there is at least one real root of $e^x \cos x + 1 = 0$. Step 1: Define $g(x) = e^x \sin x - 1$ and $h(x) = e^x \cos x + 1$. Step 2: Note that $g'(x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)$. Step 3: By Rolle's theorem, between any two roots $a$ and $b$ of $g(x)$, there exists $c \in (a,b)$ such that $g'(c) = 0$. Step 4: Set $g'(c) = 0$: $$e^c (\sin c + \cos c) = 0 \Rightarrow \sin c + \cos c = 0$$ Step 5: Rewrite $\sin c + \cos c = 0$ as $\cos c = -\sin c$. Step 6: Substitute into $h(c)$: $$h(c) = e^c \cos c + 1 = e^c (-\sin c) + 1 = -e^c \sin c + 1$$ Step 7: Since $g(c) = e^c \sin c - 1$, rearranged as $e^c \sin c = 1$, then $$h(c) = -1 + 1 = 0$$ Thus, $h(c) = 0$ at some $c$ between any two roots of $g(x)$. 4. Show $f(x) = \sin x$ satisfies the Mean Value Theorem on $[0, \pi]$. Step 1: $f$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$. Step 2: Calculate average rate of change: $$\frac{f(\pi) - f(0)}{\pi - 0} = \frac{0 - 0}{\pi} = 0$$ Step 3: Find $c$ such that $f'(c) = 0$. $$f'(x) = \cos x$$ Solve $\cos c = 0$ on $(0, \pi)$. $$c = \frac{\pi}{2}$$ Thus, MVT holds with $c = \frac{\pi}{2}$. 5. For $f(x) = \sqrt{x}$ on $[0,4]$, prove there exists $c \in (0,4)$ such that $f'(c) = \frac{1}{2\sqrt{c}}$. Step 1: $f$ is continuous on $[0,4]$ and differentiable on $(0,4)$. Step 2: Calculate average rate of change: $$\frac{f(4) - f(0)}{4 - 0} = \frac{2 - 0}{4} = \frac{1}{2}$$ Step 3: Find $f'(x)$. $$f'(x) = \frac{1}{2\sqrt{x}}$$ Step 4: By MVT, there exists $c$ such that $$f'(c) = \frac{1}{2}$$ Step 5: Solve $$\frac{1}{2\sqrt{c}} = \frac{1}{2} \Rightarrow \frac{1}{\sqrt{c}} = 1 \Rightarrow \sqrt{c} = 1 \Rightarrow c = 1$$ Since $1 \in (0,4)$, the statement is proved.