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.

$f\left(x\right)=x^5-3x^4-2x^3+6x^2+x+2;\left[1.7,1.8\right]$

Short answer

It is shown that the function has a zero in the interval$\left[1.7,1.8\right]$ and the zero is approximated to be $x=1.73$.

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(1.7) and f(1.8)

For the given function $f\left(x\right)=x^5-3x^4-2x^3+6x^2+x+2$, the function value $f(1.7)$is calculated as

$$\begin{gathered} f(1.7)={(1.7)}^5-3{(1.7)}^4-2{(1.7)}^3+6{(1.7)}^2+1.7+2 \\ f(1.7)=14.19857-25.0563-9.826+17.34+1.7+2 \\ f(1.7)=0.35627 \end{gathered}$$

And the function value $f(1.8)$is given as

$$\begin{gathered} f(1.8)={(1.8)}^5-3{(1.8)}^4-2{(1.8)}^3+6{(1.8)}^2+1.8+2 \\ f(1.8)=18.89568-31.4928-11.664+19.44+1.8+2 \\ f(1.8)=-1.02112 \end{gathered}$$

Step 3. Use Intermediate Value Theorem  

For the interval $\left[1.7,1.8\right]$, we have $1.7<1.8$and the function values $f(1.7)$and $f(1.8)$are of opposite signs. So the function has at least one zero in the interval $\left[1.7,1.8\right]$.

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 $\left[1.7,1.8\right]$the value of $x$for which the function value is $0$is $x=1.73$.

So the zero of the given function in the given interval is $x=1.73$.