Question

Use the Intermediate Value Theorem to show that each function has a zero in the given interval. Approximate the zero rounded to two decimal places.

$f\left(x\right)=x^4+8x^3-x^2+2;\left[-1,0\right]$

Short answer

It is shown that the function has a zero in the interval $\left[-1,0\right]$and the zero is approximated to berole="math" localid="1646200897947" $x=-0.6$.

Step-by-step solution

Step 1. Concept used 

The Intermediate Value Theorem states that 'Letf denote a continuous function. If $a<b$and if $f\left(a\right)$and $f\left(b\right)$are of opposite sign, then f has at least one zero betweena andb.

Step 2. Find f(-1) and f(0)

For the given function $f\left(x\right)=x^4+8x^3-x^2+2$, the function value $f(-1)$is calculated as

role="math" localid="1646201043075" $$\begin{gathered} f(-1)={(-1)}^4+8{(-1)}^3-{(-1)}^2+2 \\ f(-1)=1-8-1+2 \\ f(-1)=-6 \end{gathered}$$

And the function value $f\left(0\right)$is given as

$$\begin{gathered} f\left(0\right)=0^4+8·0^3-0^2+2 \\ f\left(0\right)=2 \end{gathered}$$

Step 3. Use Intermediate Value Theorem 

For the interval $\left[-1,0\right]$, we have $-1<0$and the function values $f(-1)$and $f\left(0\right)$are of opposite signs. So the function has at least one zero in the interval $\left[-1,0\right]$.

Step 4. Graph the function and approximate the zero 

Using a graphing utility the graph of the function is given as

With the help of utility, it is found that in the interval $\left[-1,0\right]$the value of $x$for which the function value is $0$is $x=-0.6$.

So the zero of the given function in the given interval is $x=-0.6$.