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Weekly Challenge 27: Cubic Roots

Stage: 5 Short Challenge Level: Challenge Level:1

A polynomial $f(x)$ has a factor $(x-a)$ if and only if $f(a)=0$. 
 
Thus, a polynomial cutting the $x$-axis at $10, 100, 1000$ has factors $(x-10)(x-100)(x-1000)$. This defines a cubic polynomial up to a multiplicative factor.
 
Thus
$$f(x) = A(x-10)(x-100)(x-1000) = A\left(x^3 -(10+100+1000)x^2 + .. \right)\,,$$
for some constant $A$.
 
Now, a point of inflection necessarily has $f''(x) = 0$. Only the $x^3$ and $x^2$ terms of a cubic polynomial contributes to its second derivative, so there is no need to expand the polynomial in full to see that 
$$
f''(x) = 6Ax-2220A
$$
This is zero at the single point $x = \frac{2220}{6} = 370$.

Therefore the point of inflection for the cubic is at $x=370$, regardless of the choice of $A$.
 
For the second part, the polynomial must take the form
$$
f(x) = A(x-10)(x-100)(x-a)\quad \mbox{for a constant } a \mbox{ where} \quad f''(a)=0
$$
So, we need to take the second derivative to work out the constraints on $a$. I will keep the form of the factors and use the chain rule to make life simple, although you could expand the brackets first if you wish
 
$$
f''(x) = 2A\left((x-10)+(x-100)+(x-a)\right)
$$
So, 
$$
f''(a) = 2A(2a-110)=0
$$
Since $A$ cannot be zero for a cubic polynomial, we must have $a=55$.
 
The polynomial must therefore be of the form
$$
f(x) = A(x-10)(x-55)(x-100)
$$
 
Alternative, quick, method for second part:
 From the first part of the question I noticed a generalisation that the point of inflection of a cubic is found at one third of the sum of the roots $r_1+r_2+r_3$. If one of the roots is the point of inflection then
$$
r_1+a+r_2 = 3a
$$
Thus, the point of inflection which is a root is found at one-half of the sum of the other two roots.
Thus, in our special case,
$$
a = \frac{1}{2}(10+100) = 55, \mbox{ as before}.
$$
Isn't maths great!