In this lesson, we will introduce Newton's method for finding roots. First let's do a couple practice questions which may or may not have new concepts in them to you.

For the line f(x) = 3x^{3} + 2x - 1, find where the tangent line at point (3,4) intersects the x-axis.

So, to be clear, we want to know where the *tangent line at the point (1,4)* intersects the x-axis. We *do not* want to know where the curve f(x) = 3x^{3} + 2x - 1 intersects the x-axis.

So, to find the x-axis intercept for a line, we must find the point where x=0. To do this we must get an equation for that line. In this case we need the equation for the tangent line at point (1 ,4).

So first calculate the equation for the tangent line to f(x) = 3x^{3} + 2x - 1 by differentiation - this gives us the equation for the slope of the tangent at any given point.

f ' (x) = (3)(3x^{3 - 1)} + 2x^{1 - 1)} - 0

= 9x^{2} + 2

We now have the equation f'(x) so that gives us the slope at a point. Input our point and we will find the slope at that point!

For (1, 4) , f'(x) = 9x^{2} + 2

f'(1) = 9(1)^{2} + 2 = 11

Now, to find the equation of the tangent line at point (1, 4), we have a point on that line and we know the slope, so we calculate the equation of the tangent using y = mx + b.

sub in: x = 1, y = 4, & m = 11 into y = mx + b

4 = (11)(1) + b

4 = 11 + b

b = -7

Use m = 11 & b = -7 to get the equation of our tangent (at point (1, 4)) line.

y = 11x - 7

Lastly, to find the x-axis intercept, plug in y = 0 as the is the y-value for any point on the x-axis.

0 = 11x - 7

11x = 7

x =

7

11

So the x-axis intercept of the tangent line at point (1, 4) of f(x) = 3x^{3} + 2x - 1 is (^{7}⁄_{11}, 0)