Chapter 4.2, Problem 99AYU

To determine

To find: Use the bisection method to approximate the zero of $f\left(x\right)=8x^4-2x^2+5x-1$ in the interval $\left[0,1\right]$ correct to three decimal places. Verify your result using a graphing utility.

Answer to Problem 99AYU

The value is $0.125$.

Explanation of Solution

Given:

$f\left(x\right)=8x^4-2x^2+5x-1$ and the interval is $\left[0,1\right]$

Let $a=0$ then $a+1=1$.

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

Let us compute $f\left(0\right)=8\left(0\right)-2\left(0\right)+5\left(0\right)-1=-1$.

$f\left(1\right)=8-2+5-1=10$

Since $f\left(0\right)$ and $f\left(1\right)$ are of opposite signs, we need to find the midpoint of 0 and 1.

The midpoint of 0 and 1 is $0.5$.

Let $m_1=0.5$.

The $f\left(0.5\right)=8{\left(0.5\right)}^4-2{\left(0.5\right)}^2+5\left(0.5\right)-1=1.5$

The value of $f\left(a\right)$ and $f\left(m_1\right)$ that are of opposite sign. Now evaluate $f$ at the midpoint $m_2$ of $a$ and $m_1$.

The midpoint $m_2=0.25$.

The $f\left(0.25\right)=8{\left(0.25\right)}^4-2{\left(0.25\right)}^2+5\left(0.25\right)-1=0.15625$

The value of $f\left(a\right)$ and $f\left(m_2\right)$ that are of opposite sign. Now evaluate $f$ at the midpoint $m_3$ of $a$ and $m_2$.

The midpoint $m_3=0.125$.

The $f\left(0.125\right)=8{\left(0.125\right)}^4-2{\left(0.125\right)}^2+5\left(0.125\right)-1=-0.40429$

We stop the iteration here and the value is $0.125$.