Does anyone know how to solve this?

First, find the x-intercept of g(x) = x³ - 1. x³ = 1, so x = 1. The x-intercept is (1,0).

We must now find the equation of the tangent line to the graph at (1,0):

slope of tangent line = 3x^2 with x = 1, or m = 3(1)^2 = 3.

Last, find the equation of the line with slope 3 that passes through (1,0):

0 = 3(1) + b. Then b = -3, and the equation of the tangent line at (1,0) is

y = 3x - 3.

You can find the equation of a line if you know the slope of the line and one point on the line using the point-slope form of the equation of a line.

First, let's find a point on the line. We will find the x-intercept. At the x-intercept, y = 0, so let y = 0 and solve for x.

g(x) = x^3 - 1

x^3 - 1 = 0

x^3 = 1

x = 1

The value of x at the x-intercept is 1, so the x-intercept is point (1, 0). We have the point we need.

Now we need the slope. The slope of this function at any point x is given by 3x^2. Let's find the slope at the x-intercept. At the x-intercept, x = 1, so the slope is

m = 3x^2 = 3(1)^2 = 3

We know the tangent line at the x-intercept has slope 3 and passes through the point (1, 0). Now we use the point-slope form of the equation of a line to find the equation of the tangent.

Answer: y = 3x - 3