Chapter 4.2, Problem 73AYU

To determine

To show: Using the Intermediate value theorem, $f\left(x\right)$ has a zero in the interval $\left[0,1\right]$, approximating the zero correct to 2 decimal places.

Answer to Problem 73AYU

$\text{The zero}\text{approximating the zero correct to 2 decimal places}=0.21$.

Explanation of Solution

Given:

$f\left(x\right)=8x^4-2x^2+5x-1$

Interval $=\left[0,1\right]$

$f\left(x\right)$ is a polynomial function.

Every polynomial function is a continuous function.

$f\left(x\right)=8x^4-2x^2+5x-1$

$f\left(0\right)=-1<0$

$f\left(1\right)=8{\left(1\right)}^4-2{\left(1\right)}^2+5\left(1\right)-1=10>0$

According to Intermediate value theorem, the zero lies between 0 and 1.

Dividing the interval 0 and 1 into 10 equal subintervals and evaluating $f$ at each end point.

$f\left(0.1\right)=8{\left(0.1\right)}^4-2{\left(0.1\right)}^2+5\left(0.1\right)-1=-0.5192<0$

$f\left(0.2\right)=8{\left(0.2\right)}^4-2{\left(0.2\right)}^2+5\left(0.2\right)-1=-0.7872<0$

$f\left(0.3\right)=8{\left(0.3\right)}^4-2{\left(0.3\right)}^2+5\left(0.3\right)-1=0.3848>0$

The zero lies between $0.2$ and $0.3$.

Divide the interval $0.2$ and $0.3$ into equal subintervals and evaluate $f$ at each endpoint.

$f\left(0.21\right)=8{\left(0.21\right)}^4-2{\left(0.21\right)}^2+5\left(0.21\right)-1=-0.0227<0$

$f\left(0.22\right)=8{\left(0.22\right)}^4-2{\left(0.22\right)}^2+5\left(0.22\right)-1=0.1906>0$

Zero lies between $0.21$ and $0.22$ correct to 2 decimal places.

The zero $= 0.21$.