current attempt in progress use the intermediate value theorem and the bisection method to approximate the…

Answer:

0.20

Explanation:

Step1: Evaluate function at endpoints

$f(0)=7(0)^5 - 2(0)^2+5(0)-1=-1$ $f(1)=7(1)^5 - 2(1)^2+5(1)-1=7 - 2+5 - 1 = 9$ Since $f(0)<0$ and $f(1)>0$, by the intermediate - value theorem, there is a zero in $[0,1]$.

Step2: First bisection

The mid - point of $[0,1]$ is $c_1=\frac{0 + 1}{2}=0.5$. $f(0.5)=7(0.5)^5-2(0.5)^2 + 5(0.5)-1=7\times0.03125-2\times0.25 + 2.5-1=0.21875-0.5 + 2.5-1=1.21875$ Since $f(0)<0$ and $f(0.5)>0$, the zero is in $[0,0.5]$.

Step3: Second bisection

The mid - point of $[0,0.5]$ is $c_2=\frac{0 + 0.5}{2}=0.25$. $f(0.25)=7(0.25)^5-2(0.25)^2+5(0.25)-1=7\times0.0009765625-2\times0.0625 + 1.25-1=0.0068359375-0.125+1.25-1=0.1318359375$ Since $f(0)<0$ and $f(0.25)>0$, the zero is in $[0,0.25]$.

Step4: Third bisection

The mid - point of $[0,0.25]$ is $c_3=\frac{0+0.25}{2}=0.125$. $f(0.125)=7(0.125)^5-2(0.125)^2+5(0.125)-1=7\times0.0000305176-2\times0.015625 + 0.625-1=-0.3965377007$ Since $f(0.125)<0$ and $f(0.25)>0$, the zero is in $[0.125,0.25]$.

Step5: Fourth bisection

The mid - point of $[0.125,0.25]$ is $c_4=\frac{0.125 + 0.25}{2}=0.1875$. $f(0.1875)=7(0.1875)^5-2(0.1875)^2+5(0.1875)-1=7\times0.0002755737-2\times0.03515625+0.9375-1=-0.1693772566$ Since $f(0.1875)<0$ and $f(0.25)>0$, the zero is in $[0.1875,0.25]$.

Step6: Fifth bisection

The mid - point of $[0.1875,0.25]$ is $c_5=\frac{0.1875 + 0.25}{2}=0.21875$. $f(0.21875)=7(0.21875)^5-2(0.21875)^2+5(0.21875)-1=7\times0.0005245209-2\times0.0478515625+1.09375-1=0.0304763987$ Since $f(0.1875)<0$ and $f(0.21875)>0$, the zero is in $[0.1875,0.21875]$.

Step7: Sixth bisection

The mid - point of $[0.1875,0.21875]$ is $c_6=\frac{0.1875 + 0.21875}{2}=0.203125$. $f(0.203125)=7(0.203125)^5-2(0.203125)^2+5(0.203125)-1=7\times0.0003479004-2\times0.0412609863+1.015625-1=-0.0694777425$ Since $f(0.203125)<0$ and $f(0.21875)>0$, the zero is in $[0.203125,0.21875]$. Rounding to two decimal places, we get $0.20$.