(a) Prove that if $f$ is a twice differentiable function with $f(0) = 0$ and $f(1) = 1$ and $f'(0) = f'(1) = 0$, then $|f''(x)| \geq 4$ for some $x$ in $[0,1]$.

Hint: Prove that either $f''(x) > 4$ for some $x$ in $[0,\frac{1}{2}]$, or else $f''(x) < -4$ for some $x$ in $[\frac{1}{2},1]$.

(b) Show that in fact we must have $|f''(x)| > 4$ for some $x$ in $[0,1]$.

This question is from Spivak's Calculus, Chapter 11 question 39. There is duplicate question but it seems to use integration which is not taught by Chapter 11 (on derivatives and related theorems). It also seems the Answers book for Spivak's calculus has an error, or not is explained properly, it says: "$ f(x)<4x^2$ implies $ f(1/2)<1/2 $".

  • 1
    $\begingroup$ It should be $\;f(x)<2x^2$ implies… This results from Taylor-Lagrange's inequality if $|f''(x)|<4$. $\endgroup$
    – Bernard
    Sep 15 '17 at 13:21
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    $\begingroup$ The answers book should indeed read "$f(x) < 2x^2$ implies $f(1/2) < 1/2$", which is not hard to verify by inserting $x=1/2$ and you should note that $2x^2$ has the critical second derivative of $4$. To solve (a) without integration you just need to apply the mean value theorem twice. First to show that $f''(x) < 4$ implies $f'(x) < 4x$ and then that this implies $f(x) < 2x^2$ Part (b) should then be clear by closely looking at the argument for (a). $\endgroup$
    – mlk
    Sep 15 '17 at 13:32
  • $\begingroup$ @Bernard Sorry what is Taylor-Lagrange's inequality? Can't seem to find it, $\endgroup$
    – helios321
    Sep 15 '17 at 15:12
  • $\begingroup$ @mlk How does the second application of mean value theorem give you $f(x)<2x^2$ though? $\endgroup$
    – helios321
    Sep 15 '17 at 15:35
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    $\begingroup$ Consider $g(x) = f(x)-2x^2$. Then $g'(x) =f'(x)-4x < 0$. From this, per mean value theorem $g(x) <0$ and thus $f(x) < 2x^2$. $\endgroup$
    – mlk
    Sep 15 '17 at 16:18

If $|f''(x)|<4$ for all $x\in [0,1]$ then

(1). There exists $y\in [0,1/2]$ such that $f(1/2)=f(0)+(1/2)f'(0)+(1/2)^2f''(y)/2=(1/2)^2f''(y)/2=f''(y)/8 <4/8=1/2.$


(2). There exists $z\in [1/2,1]$ such that $f(1/2)=f(1)+(1/2-1)f'(1)+(1/2-1)^2f''(z)/2= 1+f''(z)/8>1-4/8=1/2.$

  • $\begingroup$ What theorem is used to get your equation $f(1/2)=f(0)+(1/2)f'(0)+(1/2)^2f''(y)/2=(1/2)^2f''(y)/2=f''(y)/8 <4/8=1/2$? $\endgroup$
    – helios321
    Sep 16 '17 at 5:25
  • $\begingroup$ The generalized MVT, also known as a remainder formula for a power series. It should be in the book. I assume it has a section on power series. $\endgroup$ Sep 16 '17 at 19:25

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