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

  • 2
    $\begingroup$ $g'(x) \ge f'(x)$ tells you that $g(x)$ always grows at least as fast as $f(x)$ and $g(0) = f(0)$ so $g(4) \ge f(4).$ $\endgroup$
    – Doug M
    Oct 19 '20 at 8:27

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*}


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