# Finding minimum value of $g(4)$

Let $$g$$ be differentiable on $$[0, \infty)$$ with $$g(x) \geq 0$$ for all $$x \geq 0$$ with $$g(0) = 0$$. Moreover, suppose $$g'(x) \geq f'(x)$$ for all $$x\geq 0$$ where $$f(x) = x^2$$. Also suppose $$f$$ and $$g$$ are distinct functions.

How can I find the minimum possible value of $$g(4)$$?

I know we require $$g'(x) \geq 2x$$ for all $$x \geq 0$$. I'm thinking that this has to do with integration. I'm not sure how integration works with bounds because I don't think we can just integrate both sides of $$g'(x) \geq f'(x)$$ to get $$g(x) \geq f(x)$$ (or can we?). If this is allowed, then the problem's really easy: the answer would just be $$g(4) \geq f(4) = 16$$.

Any help is appreciated

• $g'(x) \ge f'(x)$ tells you that $g(x)$ always grows at least as fast as $f(x)$ and $g(0) = f(0)$ so $g(4) \ge f(4).$ Oct 19 '20 at 8:27

## 1 Answer

To answer the second part of your question: generally you cannot 'integrate' both sides of the two functions $$g'(x) \ge f'(x)$$ to obtain $$g(x) \ge f(x)$$, unless both $$f, g$$ are continuous functions for all $$x \ge 0$$ (which happens to be the case for this example). The proof of this may be achieved through the use of the Mean Value Theorem.

Thus, it suffices to conclude that

$$g'(x) \ge f'(x) \Rightarrow g(x) \ge f(x)$$

for all $$x \ge 0$$, and

\begin{align*} g(4) &\ge f(4) \\ &= 16 \end{align*}