1 function {(2,2),(4,4),(6,6),(8,8)} written is a function the other following coordinates are not functions due to it not being a straight line as a 'proper' function should be.
1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this
2) Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.
We can rewrite it as:
As for
3) Rewriting and calculating its derivative. Remember to do it, in radians.
For the second root, let's try -1.5
For x=-3.9, last root.
5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.
The generic equation for a linear function can be expressed in the slope intercept form f(x) = mx + b, where m is the slope and b is the y intercept. For this problem we can first find the equation of the line. Then we substitute x = 7 to get the f(x) value, which is n at the point x = 7.
To find the equation of the linear function we first find the slope. Slope is defined as the change in f(x) divided by the change in x. As we are given a linear function, the slope at every point is the same. We can pick any two points known to find the slope.
Let's pick (3, 7) and (9, 16). The slope m is m = (16-7)/(9-3) = 9/6 = 3/2.
Now that we have the slope, we can plug in the slope and one of the points to find b. Let's use the point (3, 7). f(x) = mx + b 7 = (1/2)(3) + b b = 11/2
Now we can write the equation f(x) = (1/2)x + 11/2