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(D) Define a new function $\phi(x) = f(x) - g(x)$.Since $f(x)$ and $g(x)$ are differentiable for all $x \ge x_0$,$\phi(x)$ is also differentiable for

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$f(x) > g(x)$ for all $x > x_0$

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(D) Define a new function $\phi(x) = f(x) - g(x)$.Since $f(x)$ and $g(x)$ are differentiable for all $x \ge x_0$,$\phi(x)$ is also differentiable for all $x \ge x_0$.The derivative is $\phi'(x) = f'(x) - g'(x)$.Given that $f'(x) > g'(x)$ for all $x > x_0$,it follows that $\phi'(x) > 0$ for all $x > x_0$.By the Mean Value Theorem,for any $x > x_0$,there exists a $c \in (x_0, x)$ such that $\frac{\phi(x) - \phi(x_0)}{x - x_0} = \phi'(c)$.Since $\phi'(c) > 0$ and $x - x_0 > 0$,we have $\phi(x) - \phi(x_0) > 0$,which implies $\phi(x) > \phi(x_0)$.Given $f(x_0) = g(x_0)$,we have $\phi(x_0) = f(x_0) - g(x_0) = 0$.Therefore,$\phi(x) > 0$ for all $x > x_0$,which means $f(x) - g(x) > 0$,or $f(x) > g(x)$ for all $x > x_0$.

Let $f(x)$ and $g(x)$ be two functions which are defined and differentiable for all $x \ge x_0$. If $f(x_0) = g(x_0)$ and $f'(x) > g'(x)$ for all $x > x_0$,then:

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