Suppose you have the function:

$f(x)=\frac{\frac{24x^3+24x}{(1+x^2)^4}}{4!}x^4$

And you want to find the value of the function in an interval $x \in [-1/2, 1/2]$.

Now, I know that the value in the outerpoints are as following:

$f(\frac{1}{2})=\frac{10}{81}$

$f(-\frac{1}{2})=-\frac{10}{81}$

My question is now:

How can I prove that this function is growing over that specific interval?

  • Simplify your expression of $f(x)$. Find its derivative, simplify it and you'll see that $f'(x)>=0$ in the interval under consideration. Hence, $f$ is "growing over that specific interval" – Bernard Massé Mar 6 '17 at 20:53
  • $f(\pm 1/2) = \pm 2/125$ if I'm not mistaken. Did you write it down wrong? – eyeballfrog Mar 6 '17 at 22:34

\begin{align}f(x)&=\frac{\frac{24x^3+24x}{(1+x^2)^4}}{4!}x^4\\ &=x^5(1+x^2)^{-3}\end{align}

\begin{align}f'(x)&=5x^4(1+x^2)^{-3}-3x^5(1+x^2)^{-4}(2x) \\&=5x^4(1+x^2)^{-3}-6x^6(1+x^2)^{-4} \\&=x^4(1+x^2)^{-4}(5(1+x^2)-6x^2) \\&=x^4(1+x^2)^{-4}(5-x^2)\end{align}

which is nonnegative over $[-\frac12,\frac12]$.

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