3-please help with highlighted portion, unsure what value goes in the box ,t= 0, 1, 2,.7+x(a) Use the stability criterion to characterize the stability of the equilibria of x4 , 1 =2(b) Use cobwebbing to find the limit that x, converges to as t→o if (i) x, = 0.5 and (ii) x, = 2.O A.В.C.OD.-18-8-8-8-0-0-82When x, = 2, X4 converges toas t→0o.Foo

The given recursion is as follows: xt+1=8xt27+xt2 The equation will be: x=8x27+x2x7+x2=8x2xx2-8x+7=0xx2-7x-x+7=0xx-7x-1=0x=0,1,7 So, the equilibrium points are x*=0,1,7Calculate the derivative of the function f(x) : f'(x)=ddx8x27+x2=16x7+x2-8x22x7+x22=112x+16x3-16x37+x22=112x7+x22 Therefore, f'(x)=112x7+x22 To determine the stability of the equilibrium value x*=0, substitute x=0 into the derivative f'(x) determined above: f'(x)=112(0)7+022=0 Therefore, f'(0)=0. So because f'(0)<1 , the stability criterion implies that: The equilibrium value x*=0 is locally stable. To determine the stability of the equilibrium value x*=1, substitute x=1 into the derivative f'(x) determined above: f'(x)=112(1)7+122=16 Therefore, f'(0)=16. So because f'(0)>1 , the stability criterion implies that: The equilibrium value x*=1 is locally unstable. To determine the stability of the equilibrium value x*=7, substitute x=7 into the derivative f'(x) determined above: f'(x)=112(7)7+722=0.25 Therefore, f'(0)=0.25. So because f'(0)<1 , the stability criterion implies that: The equilibrium value x*=7 is locally stable. CONCLUSION: The equilibrium points x=0,7 are stable and x=1 is unstable.First consider case (i), which is x0=0.5. The goal is to use cobwebbing to decide what value xt converges to as t→∞. To accomplish this, we sketch two graphs on the xt-xt+1 plane. The first graph is the graph of the recursion: xt+1=8xt27+xt2 Because x0=0.5 lies between the two equilibrium values x*=0 and x*=1, it is plausible that xt converges to one of the these two values as t→∞. Therefore when sketching the graph of f(x)=8xt27+xt2 To sketch it for 0≤xt≤1 for this part of the problem. Now plot these points in the xt-xt+1 plane and connect them with a smooth curve. Next, add the graph of the equation xt-xt+1 onto the previous graph. (This would be the line y=x if graphing on the x-y plane.To begin the cobwebbing procedure, mark x0=0.5 on the previous graph. Then, to find x1, use the fact that x1=8x027+x02 Therefore, the value x1 can be found by finding the height of the graph of f(xt) above the point x0. Mark this value on the vertical axis. Because the line xt+1=xt plays the role of the line y=x, the value x1 can be identified on the horizontal axis by drawing a vertical line where the height x1 crosses the line xt+1=xt. This forms the next two segments of the “cobweb” between the two curves xt+1=f(xt) and xt+1=xt.GRAPH: The result for the graph obtained after cobwebbing is as follows: CONCLUSION: If we repeat the procedure, we can see from the above graph that the values x0, x1, x2, x3, ............. will continue to move to the left, getting closer and closer to 0. Therefore, we can conclude that when x0=0.5, the values xt converge to the equilibrium value x*=0 as t→∞.Next consider case (ii), which is x0=2. The goal is to use cobwebbing to decide what value xt converges to as t→∞. To accomplish this, again we sketch two graphs on the xt-xt+1 plane. The first graph is again the graph of the recursion: f(xt)=8xt27+xt2 Because x0=2 lies between the two equilibrium values x*=1 and x*=7, it is plausible that xt converges to one of the these two values as t→∞. Therefore when sketching the graph of f(xt)=8xt27+xt2 To sketch it for 0≤xt≤7 for this part of the problem. Now plot these points in the xt-xt+1 plane and connect them with a smooth curve. Next, add the graph of the equation xt+1=xt onto the previous graph. (Again, this would be the line y=x if graphing on the x-y plane).To begin the cobwebbing procedure, mark x0=2 on the previous graph. Then, to find x1, use the fact that x1=8x027+x02 Therefore, the value x1 can be found by finding the height of the graph of f(xt) above the point x0. Mark this value on the vertical axis. Because the line xt+1=xt plays the role of the line y=x, the value x1 can be identified on the horizontal axis by drawing a vertical line where the height x1 crosses the line xt+1=xt. This forms the next two segments of the cobweb.GRAPH: The graph of the cobweb is as follows: CONCLUSION: If we repeat the procedure, we can see from the above graph that the values x0 , x1, x2, x3, ....... will continue to move to the right, getting closer and closer to the equilibrium value x*=7.Therefore, we can conclude that, when x0=2, the values xt converge to the equilibrium value x*=7 as t→∞.