# Help with this problem - Proof with first and second derivatives

1. May 6, 2014

### emjay66

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 $|f''(x)| \leq m$ for each $x$ in the interval $[0,a]$, and assume that $f$ takes on its largest value at an interior point of this interval. Show that $|f'(0)|+|f'(a)| \leq am$. You may assume that $f''$ is continuous on $[0,a]$

2. Relevant equations
N/A


3. The attempt at a solution
I first observed that using the Mean Value Theorem for integrals and letting $c$ be a number in the interval $[0,a]$, I can obtain
$\int_0^a|f''(t)|\,dt = |f''(c)|.(a-0) \leq m.a$
I also observed that, using the Fundamental Theorem of Calculus, I can also obtain
$\int_0^a|f''(t)|\,dt = |f'(a)| - |f'(0)|$
which would imply
$|f'(a)| - |f'(0)| \leq m.a$
I know that
$0\leq|f'(a)| - |f'(0)| \leq |f'(a) - f'(0)| \leq |f'(a)| + |f'(0)|$
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.

2. May 6, 2014

### Dick

The fundamental theorem of calculus does NOT tell you that $\int_0^a|f''(t)|\,dt = |f'(a)| - |f'(0)|$. It tells you that$\int_0^c f''(t)\,dt = f'(c) - f'(0)$. There is an interior maximum at some point x=c. You haven't used that yet. Use that.

3. May 6, 2014

### emjay66

Thanks. The correct interpretation of the FTC was a useful hint, as well as f'(c) = 0 for a c in [0,a].