Help with this problem - Proof with first and second derivatives

[emjay66]

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

Homework Statement

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]

Homework Equations

N/A

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
<br /> \int_0^a|f&#039;&#039;(t)|\,dt = |f&#039;&#039;(c)|.(a-0) \leq m.a<br />
I also observed that, using the Fundamental Theorem of Calculus, I can also obtain
<br /> \int_0^a|f&#039;&#039;(t)|\,dt = |f&#039;(a)| - |f&#039;(0)|<br />
which would imply
<br /> |f&#039;(a)| - |f&#039;(0)| \leq m.a<br />
I know that
<br /> 0\leq|f&#039;(a)| - |f&#039;(0)| \leq |f&#039;(a) - f&#039;(0)| \leq |f&#039;(a)| + |f&#039;(0)|<br />
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.

 

[Dick]

Homework Statement

I'm stuck on this problem and I'm not sure what I'm missing. The problem states:
Assume that |f&#039;&#039;(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&#039;(0)|+|f&#039;(a)| \leq am. You may assume that f&#039;&#039; is continuous on [0,a]

Homework Equations

N/A

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
<br /> \int_0^a|f&#039;&#039;(t)|\,dt = |f&#039;&#039;(c)|.(a-0) \leq m.a<br />
I also observed that, using the Fundamental Theorem of Calculus, I can also obtain
<br /> \int_0^a|f&#039;&#039;(t)|\,dt = |f&#039;(a)| - |f&#039;(0)|<br />
which would imply
<br /> |f&#039;(a)| - |f&#039;(0)| \leq m.a<br />
I know that
<br /> 0\leq|f&#039;(a)| - |f&#039;(0)| \leq |f&#039;(a) - f&#039;(0)| \leq |f&#039;(a)| + |f&#039;(0)|<br />
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.

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.

 

[emjay66]

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