Chapter 4.5, Problem 115AYU

To determine

Approximate zero of $f\left(x\right)=8x^4-2x^2+5x-1$ in the interval $\left[0,1\right]$ correct to three decimal places.

Answer to Problem 115AYU

Three decimal places $f$ has a zero at $x=0.215$ .

Explanation of Solution

Given:

  $f\left(x\right)=8x^4-2x^2+5x-1$

Calculation:

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

  $f\left(0\right)=-1$ and $f\left(1\right)=10$

Since these value of opposite sign, the intermediate value theorem implies that $f$ has a zero between $0$ and $1$ .

Now evaluate $f$ at the midpoint $0.5$ .

  $f\left(0\right)$ and $f\left(0.5\right)$ are opposite signs, $f$ has a zero between $0$ and $0.5$ .

Now find the midpoint $\frac{0+0.5}{2}=0.25$

Note that $f\left(0.25\right)=0.156$ which is positive.

This follows that $f$ has a zero between $0$ and $0.25$ .

  $f\left(0.125\right)=-0.404$ which is negative.

This follows that $f$ has a zero between $0.25$ and $0.125$ .

The point halfway between $0.25$ and $0.125$ is $0.1875$ .

We have $f\left(0.1875\right)=-0.123$ which is negative.

This follows that $f$ has a zero between $0.1875$ and $0.25$ .

The point halfway between $0.1875$ and $0.25$ is $0.21875$

We have $f\left(0.21875\right)=0.164$ which is positive. It follows that $f$ has a zero between $0.1875$ and $0.21875$ .

The point halfway between these numbers is $0.203125$

We have $f\left(0.203125\right)=-0.053$ which is negative.

This follows that $f$ has a zero between $0.21875$ and $0.203125$ .

The point halfway between these number is $0.2109375$

We have $f\left(0.2109375\right)=-0.18$ which is negative. It follows that $f$ has a zero between $0.2109375$ and $0.21875$ .

The point halfway between these numbers is $0.21875$

We have $f\left(0.21484375\right)=-0.001$ which is negative.

This follows that $f$ has a zero between $0.21484375$ and $0.21875$ .

The point halfway between these number is $0.21679375$

We have $f\left(0.21679375\right)=0.0076$ which is positive. It follows that $f$ has a zero between $0.21484375$ and $0.21679375$ .

The point halfway between these numbers is $0.21581875$

We have $f\left(0.21581875\right)=0.0033$ which is positive.

This follows that $f$ has a zero between $0.21581875$ and $0.21484375$ .

The point halfway between these number is $0.21533125$

We have $f\left(0.21533125\right)=0.0011$ which is positive. It follows that $f$ has a zero between $0.21533125$ and $0.214847375$ .

The point halfway between these numbers is $0.2150875$

We have $f\left(0.2150875\right)=0.000033$ which is positive.

This follows that $f$ has a zero between $0.2150875$ and $0.21484375$ .

The point halfway between these number is $0.214965625$

We have $f\left(0.214965625\right)=-0.0005$ which is negative. It follows that $f$ has a zero between $0.2150875$ and $0.214965625$ .

The point halfway between these numbers is $0.2150265875$

Here notice that last three numbers agree to three decimal places.

Therefore, to three decimal places $f$ has a zero at $x=0.215$ .

Conclusion:

Hence, to three decimal places $f$ has a zero at $x=0.215$ .