Answer
Lower bound = $-1.$
Upper bound = $1.$
Work Step by Step
In trying to find rational zeros of $f(x)=x^{4}+x^{3}-x-1$
(there are at most 4 real zeros),
we perform synthetic division and interpret the last row of the synthetic division table according to the Bounds on Zeros theorem.
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1}{\pm 1}$
Descart's rule
$f(x)=x^{4}+x^{3}-x-1$
has 1 change in signs$\Rightarrow$ 1 positive zero,
$f(-x)=x^{4}-x^{3}+x-1$
has $3$ changes in signs$\Rightarrow$ 3 or 1 negative zero,
Testing (with synthetic division ) $\displaystyle \frac{p}{q}=1$
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 1 & 0 & -1 &-1 \\\hline
& 1 & 2 & 2 & +1 \\\hline
1& 2 & 2 &1 & |\ \ 0 \end{array}$
- all the bottom entries are nonnegative, 1 is the positive zero
$\Rightarrow 1$ is an upper bound to the zeros of f.
Testing (with synthetic division ) $\displaystyle \frac{p}{q}=-1$
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 1 & 0 & -1 &-1 \\\hline
& -1 & 0 & 0 & +1 \\\hline
1& -0 & +0 &-1 & |\ \ 0 \end{array}$
- the bottom entries alternate in sign (or are 0), -1 is a zero of f,
$\Rightarrow -1$ is the lower bound to the zeros of f.
Lower bound = $-1.$
Upper bound = $1.$
