Find a number $M$, such that $|x^3-4x^2+x+1| < M$ for all $1<x<3$

Question

The question states that I cannot use calculus, so finding max/min of the function is not an accepted solution.

My attempt at solving the problem:

$|(x^3-4x^2) + (x+1)|$ $\le$ $|x^3-4x^2| + |x+1|$

$= |x^2||x-4| + |x+1|$

$ = x^2|x-4| + x+1$ (because 1 < x < 3 so x+1 will always be positive)

Plugging in 1 for x:

$ = 1(3) + 2$

$ = 5$

Plugging in 3 for x:

$ = 9(1) + 4 $

$ = 13 $

Therefore the value of M is 13 as there is no value in the interval from 1 < x <3 which is greater than the value outputted by 3.

Is this the right approach to the solution? When graphing the function I can see that 13 works as a value of M but there are other values that also work which are closer to the maximum of the function in that interval.

Answer 1

For all $1<x<3$ we have $$x^3-4x^2+x+1=x^3-4x^2+3x+1-2x=-x(3-x)(x-1)-2x+1<0.$$

Thus, we need to find the maximal value of $f(x)=-x^3+4x^2-x-1$ on $(1,3).$

We have $$f'(x)=-3x^2+8x-1,$$ which gives $$x_{max}=\frac{4+\sqrt{13}}{3}$$ and all $$M>f\left(\frac{4+\sqrt{13}}{3}\right)=\frac{13(5+2\sqrt{13})}{27}$$ is valid.