Find an equation of the line that is tangent to the graph of f and parallel to the given line.

Question

f(x) = x³ _______________ 3x-y+1=0 = y= 3x+1 = y' = 3 SLOPE

I need help to find x₁

Answer 1

The slope of the tangent line to f(x) = x³ is found by taking the derivative:

f '(x) = 3x² (using the power rule) and, since we know the line has a slope of 3 we can say:

3x² = 3

x² = 1

x = ±1

If x = 1, then f(1) = 1 so the point on the graph is (1. 1) and if the slope is 3 then, using the point-slope form we get: y - 1 = 3(x - 1)

If x = -1, then f(-1) = -1 so the point on the graph is (-1, -1) and, using m = 3 and the point-slope form we get: y - -1 = 3(x - -1) or y + 1 = 3(x + 1)