Monday, 30 July 2018

calculus - Consider the increasing, concave function $x^{0.5}$ on $[0, 1]$.

Consider the increasing, concave function:
$$ g(x) = \sqrt x, x ∈ [0, 1]. $$



Can you state a continuous function:
$$ f(x), x ∈ [0, 1] $$




such that $f(0) = 0, f(x)$ is twice continuously differentiable on $(0, 1]$ and:



$$ 0 < f'< g', f'' > |g''| $$



for all $x ∈ (0,1]$ ?



So basically I want an increasing function $f(x)$ which has a lower slope than $g(x)$ everywhere but is more convex than $g(x)$ is concave everywhere.

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