Let ϕ(x) = f(x) + f(2a − x) and f"(x) 0 for all x ∈ [0, a]. Then, ϕ (x)

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Let ϕ(x) = f(x) + f(2a − x) and f"(x) > 0 for all x ∈ [0, a]. Then, ϕ (x)

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Given: ϕ(x) = f(x) + f(2a − x) Differentiating above equation with respect to x we get, ϕ'(x) = f'(x) − f(2a − x) .....(1) Since, f''(x) > 0, f'(x) is an increasing function. Now, when x \in \left[ 0, a \right] x \leq 2a - x f'\left( x \right) \leq f\left( 2a - x \right) . . . . . \left( 2 \right) Considering equation (1) and (2) we get, ϕ'(x) $\leq$ 0 $\Rightarrow$ ϕ'(x) is decreasing in [0, a]

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