I am having several difficulties in solving an exercise that has been put to me.

The exercise says:

If $g: (-1; \infty) \rightarrow \mathbb{R}^+$ is a function that can be antiderivated, let $G$ be the antiderivative of $g$ so that $$G(0)=\frac{1}{2}.$$ Supposing that $$(g(x))^2=2G(x),$$ then prove that $g$ is differentiable and $$\forall x>-1, g(x)=x+1.$$

I needed help trying to unblock the situation because I don't know how to start.

Best regards

  • $\begingroup$ By the fundamental theorem of calculus $G(x)$ is differentiable and its derivative from $0$ to $x$ is $g(x)$. Use this in $(g(x))^2=2G(x)$ $\endgroup$ – DaifM Jan 16 at 11:21
  • 1
    $\begingroup$ Hint: $$ g(x) = \sqrt {2G(x)} \Rightarrow g'(x) = \frac{{G'(x)}}{{\sqrt {2G(x)} }} = \frac{{g(x)}}{{\sqrt {2G(x)} }} = \frac{{g(x)}}{{g(x)}} = 1. $$ $\endgroup$ – Gary Jan 16 at 11:24
  • $\begingroup$ @Gary $2gg'=2g$ derivation without resorting to square root is simpler. (at)OP: does $\mathbb R^+$ excludes zero ? If not we can only conclude by $G$ continuity that $g(x)=x+1$ in an interval $I$ containing zero and it can possibly be prolonged by constants outside. $\endgroup$ – zwim Jan 16 at 12:05
  • $\begingroup$ @zwim You are correct. I just thought that in this way it is easier to see that $g$ itself is differentiable. $\endgroup$ – Gary Jan 16 at 15:26
  • $\begingroup$ Ok. I see now how i can prove that g(x) is differentiable. $\endgroup$ – Davide Severino Jan 16 at 15:38

If $G$ is the antiderivative (which exists) of $g$ that is to say $G'=g$. We also have that $G(0)=\frac12$ and $g^2(x)=2G(x)$

Firstly, we know that: $$g(0)=\sqrt{2G(0)}=\sqrt{2(1/2)}=1$$

We can try differentiating the given equation: $$2g(x)g'(x)=2G'(x)$$ $$2g(x)g'(x)=2g(x)$$ we get this from the first statement we made. Now as long as $g\ne 0$ we can divide through: $$g'(x)=1$$ $$g(x)=x+C$$ now if we put in our condition for $g(0)$ we obtain: $$g(x)=x+1$$ Hope this helps :)


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