Choosing correct options for the function $g(x)=\bigl(f(x)\bigr)^2+\bigl(f'(x)\bigr)^2$ Let $g(x)=\bigl(f(x)\bigr)^2+\bigl(f'(x)\bigr)^2$, $g(0)=6$, where $f(x)$ is a thrice differentiable function such that $\bigl|f(x)\bigr|\leq 1, \; \forall x \in [-1,1]$. Then which of the following is/are true
$1.$ There is at least one point in each of the intervals $(-1,0)$ and $(0,1)$ where $\bigl|f'(x)\bigr|\leq2$
$2.$ There is at least one point in each of the intervals $(-1,0)$ and $(0,1)$ where $g(x)\leq 5$
$3.$ There is no point of local maxima of $g(x)$ in $(-1,1)$
$4.$ For some $c\in(-1,1)$, $g(c)\geq6$, $g'(c)=0$ and $g''(c) \leq 0$
I have proved option $1$ and $2$ using Lagrange's Mean Value Theorem. Using the result of option $2$, I also concluded that $g(c_1)=g(c_2)=5$ for some $c_1 \in(-1,0)$ and $c_2 \in(0,1)$ so we can conclude that $g'(c)=0$ for for some $c\in(-1,1)$ but still not able to correctly analyse option $3$ and $4$. Could someone please help me with this?
 A: First of all, I think this is a duplicate question. See Maxima and minima of $G(x)=(g(x))^2+(g'(x))^2$. Now since this question was left partly unanswered, I will try to shine some light on the matter (I am no expert).
Note that $[c_1,c_2] \subset (-1,1)$. By the Extreme Value Theorem each continuous function on a bounded interval has at least one minimum and at least one maximum. Since, $c_1 < 0 < c_2$ and $g(c_1) = g(c_2) = 5 < 6 = g(0)$ we know that a maximum on $[c_1,c_2]$ must lie in the interior. This implies that there must be at least one maximum in the interior of $(-1,1)$ and thus 3. is false.
For statement 4. note that at a maximum $c$, of a differentiable function $g(x)$, we have $g'(c) = 0$ and $g''(c) < 0$. These are called the first order condition and second order condition respectively. Now one can distinguish two cases:
Case I:
A maximum on $[c_1,c_2]$ is attained at $c = 0$. Then, $g(c) = g(0) = 6 \geq 6$, $g'(c) = 0$ and $g''(c) < 0$. In this case 4. is true.
Case II:
A maximum on $[c_1,c_2]$ is attained at $c \in (-1,0)\cup(0,1)$ (and not at $c = 0$). Since $g(0) = 6$ it must be that at $g(c)> 6$, otherwise we were in Case I. Now by the first and second order condition we have $g'(c) = 0$ and $g''(c) < 0$. So in this case 4. is true as well.
Hope this helps.
