Let $\varphi:\left(a,b \right)\rightarrow R$ twice derivatible with $\varphi\left(a \right)=\varphi\left(b \right)=0$ for all $x\in[a,b]$ Let $\varphi:\left(a,b \right)\rightarrow R$ twice derivatible with $\varphi\left(a \right)=\varphi\left(b \right)=0$ for all $x\in[a,b]$ and ${\varphi}^{(2)}(x)&gt;0$ Prove that $\varphi\left(x \right)&lt;0$ for all $x\in(a,b)$.

how ${\varphi}^{(2)}(x)&gt;0$ for all $x\in[a,b]$ we have ${\varphi}^{(1)}$ is strictly decreasing, by the rolle theorem, there exists $c\in(a,b)$ such that  ${\varphi}^{(1)}(c)=0$,  As ${\varphi}^{(1)}$ is strictly decreasing in $(a, b)$ and ${\varphi}^{(1)}(c)=0$, so ${\varphi}^{(1)}&lt;0$ in $[a,c)$ and ${\varphi}^{(1)}&gt;0$ in $(c,b]$.

Is this conclusion right?

As ${\varphi}^{(1)}$ is strictly decreasing in $(a, b)$ and ${\varphi}^{(1)}(c)=0$, so ${\varphi}^{(1)}&lt;0$ in $[a,c)$ and ${\varphi}^{(1)}&gt;0$ in $(c,b]$.
You can conclude like this: $\phi'&lt;0$ in $[a,c)$ implies that $\phi$ strictly decreases on $[a,c]$ for every $x\in [a,c], \phi(x)&lt;\phi(a)=0$. On the other hand, you have $\phi$ strictly increases on $[c,b]$, and if $x\in[c,b], \phi(x)&lt;\phi(b)=0$.
If $\varphi''(x)&gt;0$ $\forall x\in (a,b)$ then $\varphi'(x)$ is strictly incresing on $(a,b).$ Now:


Is it possible to have $\varphi'&gt;0?$ In such a case $\varphi$ will be sctrictly increasing and thus $\varphi(b)&gt;\varphi(a).$ Contradiction.
Is it possible to have $\varphi'&lt;0?$ In such a case $\varphi$ will be sctrictly decreasing and thus $\varphi(b)&lt;\varphi(a).$ Contradiction.


Thus, we have that there exists $c\in (a,b)$ such that $\varphi'&lt;0$ on $(a,c),$ $\varphi'(c)=0$ and $\varphi'&gt;0$ on $(c,b).$

Can you conclude?