Help with this problem -- Proof with first and second derivatives 1. The problem statement, all variables and given/known data I'm stuck on this problem and I'm not sure what I'm missing. The problem states: Assume that [itex] |f''(x)| \leq m[/itex] for each [itex]x [/itex] in the interval [itex][0,a] [/itex], and assume that [itex] f[/itex] takes on its largest value at an interior point of this interval. Show that [itex]|f'(0)|+|f'(a)| \leq am [/itex]. You may assume that [itex]f'' [/itex] is continuous on [itex] [0,a][/itex] 2. Relevant equations N/A [itex] [/itex] 3. The attempt at a solution I first observed that using the Mean Value Theorem for integrals and letting [itex]c[/itex] be a number in the interval [itex][0,a] [/itex], I can obtain [itex] \int_0^a|f''(t)|\,dt = |f''(c)|.(a-0) \leq m.a [/itex] I also observed that, using the Fundamental Theorem of Calculus, I can also obtain [itex] \int_0^a|f''(t)|\,dt = |f'(a)| - |f'(0)| [/itex] which would imply [itex] |f'(a)| - |f'(0)| \leq m.a [/itex] I know that [itex] 0\leq|f'(a)| - |f'(0)| \leq |f'(a) - f'(0)| \leq |f'(a)| + |f'(0)| [/itex] but I haven't been able to determine what the next step is. Based on the above information, I can't see how I can deduce the answer from what I have so far, so I'm clearly missing something. Any Hints would be very welcome.