Question

select the correct location on the table.
consider the following equations.
$f(x) = x^3 + 3x^2 - x - 6$
$g(x) = x^2 - 3x - 2$
approximate the solution to the equation $f(x) = g(x)$ using three iterations of successive approximation. use the graph as a starting point.
graph of two functions, one cubic (green) and one quadratic (blue), intersecting near x-values, with grid lines
what is the approximate value for the solution to the equation $f(x) = g(x)$?
table titled successive approximation intersection values with rows: 5/8 & 11/16; 3/4 & 13/16; 7/8 & 15/16

Explanation:

Step1: Rewrite the equation

First, set \( f(x)=g(x) \), so \( x^{3}+3x^{2}-x - 6=x^{2}-3x - 2 \). Rearrange to get \( x^{3}+2x^{2}+2x - 4 = 0 \). Let \( h(x)=x^{3}+2x^{2}+2x - 4 \). From the graph, we can see that the solution is near \( x = 1 \) (wait, maybe I misread the graph. Wait, the green and blue graphs: let's check the intersection. Wait, maybe the interval is between, say, \( x = 0.5 \) and \( x = 1 \)? Wait, no, let's check the initial interval. From the graph, the intersection seems to be in the interval where \( x \) is between, maybe, \( 0.5 \) and \( 1 \)? Wait, maybe the starting interval is \( [0,1] \)? Wait, let's check \( h(0)=-4 \), \( h(1)=1 + 2+2 - 4 = 1 \). So by Intermediate Value Theorem, there's a root in \( (0,1) \). Let's take the starting interval \( [a_0,b_0]=[0,1] \), since \( h(0)<0 \) and \( h(1)>0 \).

Step2: First iteration (n=1)

The first approximation is \( x_1=a_0+\frac{b_0 - a_0}{2}=0.5 \). Compute \( h(0.5)=(0.125)+2*(0.25)+2*(0.5)-4=0.125 + 0.5+1 - 4=-2.375<0 \). So the root is in \( [0.5,1] \), since \( h(0.5)<0 \) and \( h(1)>0 \).

Step3: Second iteration (n=2)

\( x_2=0.5+\frac{1 - 0.5}{2}=0.75 \). Compute \( h(0.75)=(0.421875)+2*(0.5625)+2*(0.75)-4=0.421875 + 1.125+1.5 - 4=-1.953125<0 \). Wait, that's still negative. Wait, maybe I made a mistake in the function. Wait, original equation: \( f(x)=x^3 + 3x^2 -x -6 \), \( g(x)=x^2 - 3x -2 \). So \( f(x)-g(x)=x^3 + 2x^2 + 2x -4 \). Wait, \( h(0.75)=0.75^3 + 2*(0.75)^2 + 2*(0.75)-4=0.421875 + 1.125 + 1.5 - 4= -1.953125 \). \( h(1)=1 + 2 + 2 -4=1 \). So root in \( [0.75,1] \).

Step4: Third iteration (n=3)

\( x_3=0.75+\frac{1 - 0.75}{2}=0.875 \) (which is \( \frac{7}{8} \)). Compute \( h(0.875)=(0.875)^3 + 2*(0.875)^2 + 2*(0.875)-4 \). \( 0.875^3=0.669921875 \), \( 2*(0.875)^2=2*0.765625=1.53125 \), \( 2*0.875=1.75 \). Sum: \( 0.669921875 + 1.53125 + 1.75 - 4=3.951171875 - 4=-0.048828125\approx -0.049<0 \). \( h(1)=1>0 \), so root in \( [0.875,1] \). Wait, but the table has \( \frac{7}{8}=0.875 \), \( \frac{15}{16}=0.9375 \), etc. Wait, maybe my initial interval was wrong. Wait, maybe the intersection is in the negative side? Wait, looking at the graph, the green graph (maybe \( f(x) \)) and blue ( \( g(x) \)): at \( x=-2 \), they intersect? Wait, no, the graph shows at \( x=-2 \), maybe? Wait, let's check \( f(-2)=(-8)+3*(4)-(-2)-6=-8 + 12 + 2 -6=0 \). \( g(-2)=4 - (-6)-2=8 \). No, that's not equal. Wait, maybe the other intersection. Wait, the green graph is \( f(x)=x^3 + 3x^2 -x -6 \), which factors? Let's factor \( f(x) \): \( x^3 + 3x^2 -x -6 \). Try \( x=1 \): \( 1 + 3 -1 -6=-3 \). \( x=2 \): \( 8 + 12 -2 -6=12 \). \( x=-2 \): \( -8 + 12 + 2 -6=0 \). Oh! \( f(-2)=0 \). \( g(-2)=4 - (-6)-2=8 \). No. \( x=-3 \): \( f(-3)=-27 + 27 + 3 -6=-3 \). \( g(-3)=9 - (-9)-2=16 \). Wait, maybe the intersection is in the positive side. Wait, maybe I messed up the function. Let's re-express \( f(x)=g(x) \): \( x^3 + 3x^2 -x -6 = x^2 - 3x -2 \) → \( x^3 + 2x^2 + 2x -4 = 0 \). Let's try \( x=1 \): \( 1 + 2 + 2 -4=1 \). \( x=0.8 \): \( 0.512 + 2*0.64 + 1.6 -4=0.512 + 1.28 + 1.6 -4=3.392 -4=-0.608 \). \( x=0.9 \): \( 0.729 + 2*0.81 + 1.8 -4=0.729 + 1.62 + 1.8 -4=4.149 -4=0.149 \). So root between 0.8 and 0.9. \( x=0.875 \) (7/8=0.875): \( f(0.875)-g(0.875)= (0.875)^3 + 2*(0.875)^2 + 2*(0.875)-4 \approx 0.67 + 1.53 + 1.75 -4=3.95 -4=-0.05 \). \( x=0.875 \) is -0.05, \( x=0.9 \) is 0.15. So the root is near 0.875 (7/8) or 15/16 (0.9375)? Wait, the table has 7/8, 15/16, etc. Wait, maybe the starting interval was [0,1], first iteration x=0.5, second x=0.75 (3/4), third x=0.875 (7/8). Wait, but when we computed h(0.875)≈-0.05, which is close to zero. So the approximate value is 7/8? Wait, but let's check the table. The table has 5/8, 11/16, 3/4, 13/16, 7/8, 15/16. 7/8 is 0.875, which we calculated. So the correct location is 7/8 (the cell with 7/8 in the left column, maybe? Wait, the table is:

Row 1: 5/8, 11/16

Row 2: 3/4, 13/16

Row 3: 7/8, 15/16

So after three iterations, the approximation is 7/8 (0.875), since at x=7/8, h(x)≈-0.05, which is close to zero, and the next iteration would be 15/16, but we did three iterations. Wait, maybe my initial interval was wrong. Wait, maybe the intersection is in the negative side? Let's check x=-2: f(-2)=0, g(-2)=4 - (-6)-2=8, not equal. x=-3: f(-3)=-27 + 27 + 3 -6=-3, g(-3)=9 + 9 -2=16. No. x=1: f(1)=1 + 3 -1 -6=-3, g(1)=1 - 3 -2=-4. Wait, f(1)-g(1)=1. Oh! Wait, I made a mistake earlier. f(1)=1 + 3 -1 -6=-3, g(1)=1 - 3 -2=-4. So f(1)-g(1)=1. f(0)=0 + 0 -0 -6=-6, g(0)=0 -0 -2=-2. So f(0)-g(0)=-4. So root between 0 and 1, since f(0)-g(0)=-4 <0, f(1)-g(1)=1>0. So h(x)=f(x)-g(x)=x^3 + 2x^2 + 2x -4. Wait, no: f(x)-g(x)=x^3 + 3x^2 -x -6 - (x^2 - 3x -2)=x^3 + 2x^2 + 2x -4. Yes. So h(0)=-4, h(1)=1. First iteration: x1=0.5, h(0.5)=-2.375. Second: x2=0.75, h(0.75)=-1.953. Third: x3=0.875, h(0.875)=-0.048. So after three iterations, x3=0.875=7/8. So the approximate value is 7/8, which is in the table's third row, first column (7/8).

Answer:

The approximate value for the solution to the equation \( f(x) = g(x) \) is \(\frac{7}{8}\) (the cell with \(\frac{7}{8}\) in the table, specifically the left - most cell of the third row in the "Successive Approximation Intersection Values" table).