# Problem Of The Week #424 July 6th, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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Given that $f(x)=ax^3+bx^2+cx+d$ and $|f'(x)|\le 1$ for $0\le x \le 1$. Find the maximum value of $a$.

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#### anemone

##### MHB POTW Director
Staff member
Congratulations to castor28 for his correct solution , which you can find below:
We must fit the parabola $y=f'(x)=3x^2+2bx+c$ within the rectangle $[0,1]\times[-1,1]$.

To get the largest value of $a$ (the steepest parabola), we must have $f'(0)=f'(1)=1$, the axis at $x=\dfrac12$, and $f'\left(\dfrac12\right)=-1$:

This gives $f'(x)= 8\left(x-\dfrac12\right)^2-1= 8x^2-8x+1$ and $\bf a=\dfrac83$.

More precisely, the condition will be satisfied for $\lvert a\rvert\le\dfrac83$.

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