How many real roots of $f(x)=x^{12}-x^9+x^4-x+1$ between $0$ and $1$?

Question

How can we know if $f(x) = x^{12} - x^9 + x^4 - x + 1$ has How many real roots between 0 and 1. ?

Okay so at both 1 and 0 the value of function is 1. So there can be no or even roots between them so how can we tell can we use calculus ?

Answer 1

If $0<x<1$ then $$f(x)=1-x+x^4-x^9+x^{12}=1-x+x^4(1-x^5)+x^{12}>0$$

Answer 2

$\begin{cases} f(x)=x^{12}-x^9+x^4-x+1 \\ f'(x)=12x^{11}-9x^8+4x^3-1 \\ f''(x)=132x^{10}-72x^7+12x^2=12x^2(11x^8-6x^5+1)=12x^2g(x) \\ g'(x)=88x^7-30x^4=x^4(88x^3-30) \end{cases}$

And we can stop here, because we are only interested in the signs of the derivatives.

$a=\sqrt[3]{\frac{30}{88}}$

$\begin{array}{|c|ccccc|} \hline x & 0 && a && 1\\ \hline g'& 0 & - & 0 & + & 58 \\ \hline g & 1 & \searrow & \simeq 0.62 & \nearrow & 6 \\ \hline f'' & 0 && + && 72\\ \hline f' & -1 && \nearrow && 6 \\ \hline\end{array}$

So there exists $b\in]0,a[$ such that $f'(b)=0$.

$\begin{array}{|c|ccccc|} \hline x & 0 && b && 1\\ \hline f'& -1 & - & 0 & + & 6 \\ \hline f & 1 & \searrow & f(b) & \nearrow & 1 \\ \hline \hline\end{array}$

From there you can calculate $b$ by dichotomy $b\simeq 0.6746$ and then compute $f(b)\simeq 0.51245>0$. We are dealing with polynomial here, so continuity is assured and abrupt variations excluded so we can be quite confident that $f>0$ on $[0,1]$.

Now, if you want to pursue the algebraic proof, you'll have to substitute $b$ : we know that $bf'(b)=12b^{12}-9b^9+4b^4-b=0$

So $12f(b)=(9b^9-4b^4+b)+12(-b^9+b^4-b+1)=-3b^9+8b^4-11b+12$

And then carry on with this lesser degree equation until you get $f(b)>0$.

You wanted a general method without tricks, you got one.