Let f : [0, 2] [ ] R be a twice differentiable function such that f''(x) 0, for all x [ …

Question

Let f : [0, 2] $$ \to $$ R be a twice differentiable
function such that f''(x) > 0, for all x $$ \in $$ (0, 2).
If $$\phi (x) = f(x) + f(2 -- x), then \[\phi $$ is :

Accepted Answer

$$\phi (x) = f(x) + f(2 -- x) \[ \Rightarrow $$ $$\phi $$'(x) = f'(x) - f'(2 -- x) Since f''(x) > 0 for all x $$ \in $$ (0, 2) $$ \Rightarrow $$ f'(x) is an increasing function for all x $$ \in $$ (0, 2). Case 1 : When $$\phi (x) is increasing function So \[\phi $$'(x) > 0 $$ \Rightarrow $$ f'(x) - f'(2 -- x) > 0 $$ \Rightarrow $$ f'(x) > f'(2 -- x) $$ \Rightarrow $$ x > 2 -- x $$ \Rightarrow $$ x > 1 $$ \therefore $$ $$\phi (x) is increasing on (1, 2). Case 2 : When \[\phi (x) is decreasing function So \[\phi $$'(x) < 0 $$ \Rightarrow $$ f'(x) - f'(2 -- x) < 0 $$ \Rightarrow $$ f'(x) < f'(2 -- x) $$ \Rightarrow $$ x < 2 -- x $$ \Rightarrow $$ x < 1 $$ \therefore $$ \[\phi (x) is decreasing on (0, 1).

Solutions · 1