F=x-e^x=0. Find the root between a=0 and b=1 by using bisection method
Answers
Answer:
We have to find the third approximation of root of the equation x
3
−x
2
−1=0 in the interval (1,2) using successive Bisection method.
Iteration 1: k=0
c
0
=
2
a
0
+b
0
=
2
1+2
=1.5
Since f(c
0
)f(a
0
)=f(1.5)f(1)<0
Therefore set a
1
=a
0
,b
1
=c
0
Iteration 2: k=1
c
1
=
2
a
1
+b
1
=
2
1+1.5
=1.25
Since f(c
1
)f(a
1
)=f(1.25)f(1)>0
Therefore set a
2
=c
1
,b
2
=b
1
Iteration 3: k=2
c
2
=
2
a
2
+b
2
=
2
1.25+1.5
=1.375
Thus the third approximation of the root is 1.375 respectively.
Answer:
i hope it helps u enjoy :)
Step-by-step explanation:
for this problem. We're trying to figure out whether this function here actually has an X intercept or zero of the function or root of the function here on this close interval from 1 to 1.5. Now, there's a couple of things that we would need Thio have is requirements. Before we can actually figure that out. We're going to be used the intermediate value theorem here and the first thing required by the Intermediate Value Theorem is that you have to have a continuous function on the interval. You're considering to notice how we have for our function f of X to the X power is the first term. Well, that's a exponential function, So that's continuous, actually everywhere. Notice how your second term negative X cubed is also continuous everywhere. It's a ah part of a polynomial here, so we definitely are continuous everywhere with each of these terms. And subtracting one from another is not gonna change that. We're adding them together. Won't change that either. You learned that back in section 2.4. So now I know that this entire function here is to the X power minus X cubed is continuous not only in the interval we're considering, but actually everywhere. So we'll be solid throughout the entire problem here with the fact that we have a continuous function, we won't have to worry about that at all. The next step in the process is checking out what's happening at these in points here at both at one and 1.5 to see what the Y values look like. If you have and we actually have it here, I've already written down. If you have, why values that have opposite signs in each of these values on the ends of the interval, along with a continuous function. Of course, that means that you have to have it least one X value in the interval from one of 1.5 at which you have an ex intercept or a zero of the function. In other words, that why value is zero. You notice how it one year Y value was one, and at 1.5 your wire function value is negative. One negative 10.547 notice other opposite signs of each other, so coupled with the fact that you have a continuous function on this interval, and why values with opposite signs on the in points. Well, that means by the intermediate value there, um, there has to exist. I'm going to race this real quick here. There has to exist at least one. So stay this again, at least one value, and I'll call the value, See? So we'll say at least one see value inside. This interval in the interval again is from one to 1.5. Such that Well, you could just simply say, where the why value or function value at this see value, you simply say F C is zero equals zero. And this would be an ex intercept for that function. So we know all of this here. The bottom by the ivy. T you satisfied all the requirements for it? Okay, well, now that we've established that we have definitely a zero between one and 1.5 and part B, we're gonna get step further. Notice how the interval length is 0.5 here from one in point to the other. What they're gonna do is narrow it down even further. Remember earlier was from 1 to 1.5. Now we're going from 1.25 to 1.5 and now the interval is just 1/4 with the part. They're with the cross. Notice how I've already gone ahead and written down The Y values each of these 10 points, but we've already established earlier that this function f of X is continuous everywhere. So we've already got a continuous function on the interval here and then just have the Y values at these in points have opposite signs. Once again, we already knew the one at 1.5, it was negative 0.547 at 1.25 The Y values positive 0.4 to 5. So the same thing is happening. That happened earlier on this interval from 1 to 1.5. We could say