Chapter 4.2, Problem 74AYU

To determine

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

Answer to Problem 74AYU

$\text{The zero}\text{approximating the zero correct to 2 decimal places}=-0.60$.

Explanation of Solution

Given:

$f\left(x\right)=x^4+8x^3-x^2+2$

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

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

Every polynomial function is a continuous function.

$f\left(x\right)=x^4+8x^3-x^2+2$

$f\left(-1\right)=-6<0$

$f\left(0\right)=2>0$

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

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

$f\left(-0.9\right)={\left(-0.9\right)}^4+8{\left(-0.9\right)}^3-{\left(0.9\right)}^2+2=-4.0569<0$

$f\left(-0.8\right)={\left(-0.8\right)}^4+8{\left(-0.8\right)}^3-{\left(0.8\right)}^2+2=-2.3264<0$

$f\left(-0.6\right)={\left(-0.6\right)}^4+8{\left(-0.6\right)}^3-{\left(0.6\right)}^2+2=-0.0884<0$

$f\left(-0.5\right)={\left(-0.5\right)}^4+8{\left(-0.5\right)}^3-{\left(0.5\right)}^2+2=1.0375>0$

$f\left(-0.2\right)={\left(-0.2\right)}^4+8{\left(-0.2\right)}^3-{\left(0.2\right)}^2+2=1.8976>0$

$f\left(-0.3\right)={\left(-0.3\right)}^4+8{\left(-0.3\right)}^3-{\left(0.3\right)}^2+2=1.7021>0$

The zero lies between $-0.6$ and $-0.5$.

Divide the interval $-0.6$ and $-0.5$ into equal subintervals and evaluate $f$ at each endpoint.

$f\left(-0.59\right)={\left(-0.59\right)}^4+8{\left(-0.59\right)}^3-{\left(0.59\right)}^2+2=0.13>0$

Zero lies between $-0.60$ and $-0.59$ correct to 2 decimal places.

The zero $=-0.60$.