Question

Find the bounds to the zeros of each polynomial function. Use the bounds to obtain a complete graph of f.

$f\left(x\right)=x^3-5x^2-11x+11$

Short answer

Every zero of the polynomial function will lie between $-12$and $12$.

The graph of the function is as follows,

Step-by-step solution

Step 1. Given Information  

We are given a polynomial function,

$f\left(x\right)=x^3-5x^2-11x+11$

We have to find the bounds to the zeros of the function and using it we need to obtain its graph.

Step 2. Concept used  

Let f denote a polynomial function whose leading coefficient is $1$.

$f\left(x\right)=x^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0$

A bound M on the real zeros of f is the smaller of the two numbers

$Max\left\{1,\left|a_0\right|+\left|a_1\right|+...+\left|a_{n-1}\right|\right\},1+Max\left\{\left|a_0\right|,\left|a_1\right|,...,\left|a_{n-1}\right|\right\}$

where Max means “choose the largest entry in .”

Step 3. Comparing the function with standard form

The leading coefficient is already one, so

On comparing with standard form we get,

$$\begin{gathered} a_2=-5 \\ {a}_1=-11 \\ {a}_0=11 \end{gathered}$$

Step 4. Find the two numbers

Using the formula, the two numbers are found as,

$$\begin{gathered} Max\left\{1,\left|a_0\right|+\left|a_1\right|+\dots +\left|a_{n-1}\right|\right\}=Max\{1,|-5|+|-11|+|11\left|\right\} \\ =Max\{1,27\} \\ =27 \end{gathered}$$

And,

$$\begin{gathered} 1+Max\left\{\left|a_0\right|,\left|a_1\right|,\dots ,\left|a_{n-1}\right|\right\}=1+Max\left\{\right|-5|,|-11|,|1|\mid \} \\ =1+11 \\ =12 \end{gathered}$$

Step 5. Finding the bounds  

Among the two numbers $27$and $12$, $12$is the smallest.

Therefore, $12$is the bound.

Every real zero of f lies between $-12$and $12$.

Step 6. Graphing the function  

Using the bound the graph of the function is as follows,