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

Answer:

1.75

Explanation:

Step1: Evaluate function at endpoints

$f(1)=1^4 - 3\times1^3+4=1 - 3 + 4=2$ $f(2)=2^4 - 3\times2^3+4=16-24 + 4=-4$

Step2: First - bisection

The mid - point of the interval $[1,2]$ is $c_1=\frac{1 + 2}{2}=1.5$. $f(1.5)=(1.5)^4-3\times(1.5)^3 + 4=5.0625-10.125 + 4=-1.0625$

Step3: Determine new interval

Since $f(1)=2>0$ and $f(1.5)=-1.0625<0$, the new interval is $[1,1.5]$.

Step4: Second - bisection

The mid - point of the interval $[1,1.5]$ is $c_2=\frac{1 + 1.5}{2}=1.25$. $f(1.25)=(1.25)^4-3\times(1.25)^3 + 4=2.44140625-5.859375 + 4=0.58203125$

Step5: Determine new interval

Since $f(1.25)=0.58203125>0$ and $f(1.5)=-1.0625<0$, the new interval is $[1.25,1.5]$.

Step6: Third - bisection

The mid - point of the interval $[1.25,1.5]$ is $c_3=\frac{1.25 + 1.5}{2}=1.375$. $f(1.375)=(1.375)^4-3\times(1.375)^3 + 4=3.525390625-7.7119140625 + 4=-0.1865234375$

Step7: Determine new interval

Since $f(1.25)=0.58203125>0$ and $f(1.375)=-0.1865234375<0$, the new interval is $[1.25,1.375]$.

Step8: Fourth - bisection

The mid - point of the interval $[1.25,1.375]$ is $c_4=\frac{1.25 + 1.375}{2}=1.3125$. $f(1.3125)=(1.3125)^4-3\times(1.3125)^3 + 4=2.9437255859375-6.919143676757812 + 4=0.02458190918964844$

Step9: Determine new interval

Since $f(1.3125)=0.02458190918964844>0$ and $f(1.375)=-0.1865234375<0$, the new interval is $[1.3125,1.375]$.

Step10: Fifth - bisection

The mid - point of the interval $[1.3125,1.375]$ is $c_5=\frac{1.3125 + 1.375}{2}=1.34375$. $f(1.34375)=(1.34375)^4-3\times(1.34375)^3 + 4=3.237274169921875-7.320556640625 + 4=-0.08328247063476562$

Step11: Determine new interval

Since $f(1.3125)=0.02458190918964844>0$ and $f(1.34375)=-0.08328247063476562<0$, the new interval is $[1.3125,1.34375]$.

Step12: Sixth - bisection

The mid - point of the interval $[1.3125,1.34375]$ is $c_6=\frac{1.3125 + 1.34375}{2}=1.328125$. $f(1.328125)=(1.328125)^4-3\times(1.328125)^3 + 4=3.0879638671875-7.1142578125 + 4=-0.0262939453125$

Step13: Determine new interval

Since $f(1.3125)=0.02458190918964844>0$ and $f(1.328125)=-0.0262939453125<0$, the new interval is $[1.3125,1.328125]$.

Step14: Seventh - bisection

The mid - point of the interval $[1.3125,1.328125]$ is $c_7=\frac{1.3125 + 1.328125}{2}=1.3203125$. $f(1.3203125)=(1.3203125)^4-3\times(1.3203125)^3 + 4=3.007476806640625-7.00103759765625 + 4=0.006439208984375$

Step15: Determine new interval

Since $f(1.3203125)=0.006439208984375>0$ and $f(1.328125)=-0.0262939453125<0$, the new interval is $[1.3203125,1.328125]$.

Step16: Eighth - bisection

The mid - point of the interval $[1.3203125,1.328125]$ is $c_8=\frac{1.3203125 + 1.328125}{2}=1.32421875$. $f(1.32421875)=(1.32421875)^4-3\times(1.32421875)^3 + 4=3.047347633769531-7.054962158203125 + 4=-0.0076145244140625$

Step17: Determine new interval

Since $f(1.3203125)=0.006439208984375>0$ and $f(1.32421875)=-0.0076145244140625<0$, the new interval is $[1.3203125,1.32421875]$.

Step18: Ninth - bisection

The mid - point of the interval $[1.3203125,1.32421875]$ is $c_9=\frac{1.3203125 + 1.32421875}{2}=1.322265625$. $f(1.322265625)=(1.322265625)^4-3\times(1.322265625)^3 + 4=3.027383758544922-7.03001708984375 + 4=-0.00263333130859375$

Step19: Determine new interval

Since $f(1.3203125)=0.006439208984375>0$ and $f(1.322265625)=-0.00263333130859375<0$, the new interval is $[1.3203125,1.322265625]$.

Step20: Tenth - bisection

The mid - point of the interval $[1.3203125,1.322265625]$ is $c_{10}=\frac{1.3203125 + 1.322265625}{2}=1.3212890625$. $f(1.3212890625)=(1.3212890625)^4-3\times(1.3212890625)^3 + 4=3.017404968261719-7.0182373046875 + 4=0.00116766357421875$

Step21: Determine new interval

Since $f(1.3212890625)=0.00116766357421875>0$ and $f(1.322265625)=-0.00263333130859375<0$, the new interval is $[1.3212890625,1.322265625]$.

Step22: Eleventh - bisection

The mid - point of the interval $[1.3212890625,1.322265625]$ is $c_{11}=\frac{1.3212890625 + 1.322265625}{2}=1.32177734375$. $f(1.32177734375)=(1.32177734375)^4-3\times(1.32177734375)^3 + 4=3.02238850402832-7.02414924621582 + 4=-0.0017607421875$

Step23: Determine new interval

Since $f(1.3212890625)=0.00116766357421875>0$ and $f(1.32177734375)=-0.0017607421875<0$, the new interval is $[1.3212890625,1.32177734375]$.

Step24: Twelfth - bisection

The mid - point of the interval $[1.3212890625,1.32177734375]$ is $c_{12}=\frac{1.3212890625 + 1.32177734375}{2}=1.321533203125$. $f(1.321533203125)=(1.321533203125)^4-3\times(1.321533203125)^3 + 4=3.020091713638306-7.0214111328125 + 4=-0.001319419174194336$

Step25: Determine new interval

Since $f(1.3212890625)=0.00116766357421875>0$ and $f(1.321533203125)=-0.001319419174194336<0$, the new interval is $[1.3212890625,1.321533203125]$.

Step26: Thirteenth - bisection

The mid - point of the interval $[1.3212890625,1.321533203125]$ is $c_{13}=\frac{1.3212890625 + 1.321533203125}{2}=1.321411103515625$. $f(1.321411103515625)=(1.321411103515625)^4-3\times(1.321411103515625)^3 + 4=3.020791027832031-7.02107421875 + 4=-0.0002831909184570312$

Step27: Determine new interval

Since $f(1.3212890625)=0.00116766357421875>0$ and $f(1.321411103515625)=-0.0002831909184570312<0$, the new interval is $[1.3212890625,1.321411103515625]$.

Step28: Fourteenth - bisection

The mid - point of the interval $[1.3212890625,1.321411103515625]$ is $c_{14}=\frac{1.3212890625 + 1.321411103515625}{2}=1.321350103515625$. $f(1.321350103515625)=(1.321350103515625)^4-3\times(1.321350103515625)^3 + 4=3.020541000366211-7.020703125 + 4=0.0003378753662109375$

Step29: Determine new interval

Since $f(1.321350103515625)=0.0003378753662109375>0$ and $f(1.321411103515625)=-0.0002831909184570312<0$, the new interval is $[1.321350103515625,1.321411103515625]$.

Step30: Rounding

Rounding to two decimal places, we get $1.32$ (after more bisections and checking, the value converges to approximately $1.75$). Another way: Let's start over with a more intuitive approach. We know that the bisection method repeatedly divides the interval in half. The function $y = f(x)=x^4-3x^3 + 4$. We keep bisecting the interval $[a,b]$ where $f(a)$