The properties listed are:

1) f is infinitely differentiable on the real numbers

2) $f(0) = 1, f'(0) = 1$, and $f''(0) = 2$

3) $|f'''(x)| < b$ for all x in $[0,1]$

This is question 42 from the GRE math subject test 9367.

My idea was to integrate both sides of the inequality in property three, and plug in the info from property 2 for the value of c (post integration). Is this a valid approach? I get the correct answer for b, which is 12, but I don't know if this is mathematically valid to integrate a magnitude, and integrate both sides of an inequality.

Please help! Thank you.

  • 2
    $\begingroup$ Your approach is correct and the justification is as follows If $f$ and $g$ are continuous functions on $[a,b]$ and $f(x) \leq g(x)$ for all $x \in [a,b]$, then $\int_\limits{a}^{b}f \leq \int_\limits{a}^{b}g$ and since you are given that $f$ is infinitely differentiable which in particular means that derivatives of all orders of $f$ are continuous, so you can apply your trick. $\endgroup$ Apr 7 '15 at 17:33
  • $\begingroup$ Thank you so much for your help. May I ask how would I justify integrating the magnitude of f'''(x)? I feel like this is the most uncertain part of my proof/solution. $\endgroup$
    – anonymous
    Apr 7 '15 at 17:36
  • 1
    $\begingroup$ you can not integrate the modulus what you should do is that you should integrate the two sides of the following inequality $-b<f(x)<b$. $\endgroup$ Apr 7 '15 at 17:40
  • $\begingroup$ Great idea, @UrbanPENDU!!! Thank you! :) $\endgroup$
    – anonymous
    Apr 7 '15 at 17:43

Taylor says $f(1)= f(0)+f'(0)\cdot 1 +f''(0)\cdot 1^2/2 +f'''(c_x)\cdot 1^3/6 = 1+1+1+f'''(c_x)/6 $ for some $c_x\in (0,1).$ Thus $f(1) \le 3 + b/6.$ This is less than $5$ if $b<12.$ If $b\ge 12,$ then the polynomial $p(x)=1+x +x^2/2 +bx^3/6$ satisfies $p(1)\ge 5.$


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