Question:

Step 1: Apply the Mean Value Theorem to construct auxiliary functions. Given $f(x)$ and $g(x)$ are differentiable functions for $0 \leq x \leq 1$ with $f(0) = 2$, $g(0) = 0$, $f(1) = 6$, and $g(1) = 2$. The problem states that $c$ may be different for each statement. For each statement, we construct a suitable function and apply the Mean Value Theorem on the interval $[0,1]$. Step 2: Evaluate the first statement: $f'(c) - f(0) = g'(c)$. This statement can be rewritten as $f'(c) - g'(c) = f(0)$. Consider the function $h(x) = f(x) - g(x)$. $h(x)$ is differentiable on $[0,1]$. $h(0) = f(0) - g(0) = 2 - 0 = 2$. $h(1) = f(1) - g(1) = 6 - 2 = 4$. By the Mean Value Theorem, there exists $c_1 \in (0,1)$ such that $$h'(c_1) = \frac{h(1) - h(0)}{1 - 0} = \frac{4 - 2}{1} = 2.$$ Since $h'(c_1) = f'(c_1) - g'(c_1)$, we have $f'(c_1) - g'(c_1) = 2$. As $f(0) = 2$, the statement $f'(c_1) - g'(c_1) = f(0)$ is true for this $c_1$. Step 3: Evaluate the second statement: $f'(c) - g(0) = 2g'(c)$. This statement can be rewritten as $f'(c) - 2g'(c) = g(0)$. Consider the function $k(x) = f(x) - 2g(x)$. $k(x)$ is differentiable on $[0,1]$. $k(0) = f(0) - 2g(0) = 2 - 2(0) = 2$. $k(1) = f(1) - 2g(1) = 6 - 2(2) = 6 - 4 = 2$. By the Mean Value Theorem, there exists $c_2 \in (0,1)$ such that $$k'(c_2) = \frac{k(1) - k(0)}{1 - 0} = \frac{2 - 2}{1} = 0.$$ Since $k'(c_2) = f'(c_2) - 2g'(c_2)$, we have $f'(c_2) - 2g'(c_2) = 0$. As $g(0) = 0$, the statement $f'(c_2) - 2g'(c_2) = g(0)$ is true for this $c_2$. Step 4: Evaluate the third statement: $f'(c) + f(1) = 3g'(c)$. This statement can be rewritten as $f'(c) - 3g'(c) = -f(1)$. Consider the function $m(x) = f(x) - 3g(x)$. $m(x)$ is differentiable on $[0,1]$. $m(0) = f(0) - 3g(0) = 2 - 3(0) = 2$. $m(1) = f(1) - 3g(1) = 6 - 3(2) = 6 - 6 = 0$. By the Mean Value Theorem, there exists $c_3 \in (0,1)$ such that $$m'(c_3) = \frac{m(1) - m(0)}{1 - 0} = \frac{0 - 2}{1} = -2.$$ Since $m'(c_3) = f'(c_3) - 3g'(c_3)$, we have $f'(c_3) - 3g'(c_3) = -2$. The statement requires $f'(c) - 3g'(c) = -f(1)$. Since $-f(1) = -6$, the statement would imply $-2 = -6$, which is false. Therefore, this statement is not true for any $c \in (0,1)$. Step 5: Evaluate the fourth statement: $f'(c) + 2g(1) = 4g'(c)$. This statement can be rewritten as $f'(c) - 4g'(c) = -2g(1)$. Consider the function $n(x) = f(x) - 4g(x)$. $n(x)$ is differentiable on $[0,1]$. $n(0) = f(0) - 4g(0) = 2 - 4(0) = 2$. $n(1) = f(1) - 4g(1) = 6 - 4(2) = 6 - 8 = -2$. By the Mean Value Theorem, there exists $c_4 \in (0,1)$ such that $$n'(c_4) = \frac{n(1) - n(0)}{1 - 0} = \frac{-2 - 2}{1} = -4.$$ Since $n'(c_4) = f'(c_4) - 4g'(c_4)$, we have $f'(c_4) - 4g'(c_4) = -4$. As $-2g(1) = -2(2) = -4$, the statement $f'(c_4) - 4g'(c_4) = -2g(1)$ is true for this $c_4$. The statements that are true for some $0 < c < 1$ are: 1. $f'(c) - f(0) = g'(c)$ 2. $f'(c) - g(0) = 2g'(c)$ 4. $f'(c) + 2g(1) = 4g'(c)$