# Given a relation between $G$ and $g$ prove that $g(x)=x+1$.

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

• By the fundamental theorem of calculus $G(x)$ is differentiable and its derivative from $0$ to $x$ is $g(x)$. Use this in $(g(x))^2=2G(x)$ – DaifM Jan 16 at 11:21
• Hint: $$g(x) = \sqrt {2G(x)} \Rightarrow g'(x) = \frac{{G'(x)}}{{\sqrt {2G(x)} }} = \frac{{g(x)}}{{\sqrt {2G(x)} }} = \frac{{g(x)}}{{g(x)}} = 1.$$ – Gary Jan 16 at 11:24
• @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. – zwim Jan 16 at 12:05
• @zwim You are correct. I just thought that in this way it is easier to see that $g$ itself is differentiable. – Gary Jan 16 at 15:26
• Ok. I see now how i can prove that g(x) is differentiable. – 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 :)