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Let $h:[0,1] \to (0,\infty)$ be a continuously differentiable function such that the following inequality is true:

$$\frac{h'(t)}{h(t)} > -\frac{1}{2} \ \ \ \ \text{for all $t\in (0,1)$} .$$ Let function $f$ be defined as

$$f(x) = \int_0^{x^2} h(t) dt.$$ a) prove that $f$ is twice continuously differentiable convex function on the interval $(0,1).$

b) prove that there exists(only one) $c \in (0,1)$ such that

$$ f'(c)= \int_0^1 h(t) dt.$$ a) Since I have to prove that $f$ is convex, I have to show that $f''(x)>0 $ for all $x \in (0,1).$ I get $f'(x)=h(x)2x.$ The second derivative is $f''(x)=h'(x)2x+2h(x).$ I get $$h'(x)2x+2h(x)>-\frac{h(x)x}{2}+2h(x) = h(x)\left(-\frac{x}{2}+2 \right)>0.$$ Is this correct? I don't know how to solve the second question.

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Hint: use Mean value $$f'(c)= f(1)-f(0)=RHS$$

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