Let f(x) = x^3 - x^2 + x - 1. Find an equation for tangent line to f at x = 2.
To find the equation for the tangent line to the curve at a given point, we need to determine both the slope of the curve at that point (also known as the derivative) and the coordinates of the point itself.
First, let's find the derivative of the function f(x). The derivative of x^n, where n is a constant, is given by n*x^(n-1). Applying this rule to each term of f(x):
f'(x) = d/dx (x^3 - x^2 + x - 1)
= 3x^2 - 2x + 1
Now we have the derivative of the function. We need to find the slope of the tangent line at x = 2, so let's substitute x = 2 into f'(x):
f'(2) = 3(2)^2 - 2(2) + 1
= 12 - 4 + 1
= 9
The slope of the tangent line at x = 2 is 9.
Next, we need to find the coordinates of the point (2, f(2)) on the curve. Substituting x = 2 into f(x):
f(2) = (2)^3 - (2)^2 + (2) - 1
= 8 - 4 + 2 - 1
= 5
Therefore, the point on the curve is (2, 5).
Now we have the slope of the tangent line (9) and a point on the line (2, 5). We can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is the point on the line. Substituting the values we found:
y - 5 = 9(x - 2)
Expanding the equation:
y - 5 = 9x - 18
Adding 5 to both sides:
y = 9x - 13
Therefore, the equation of the tangent line to f(x) at x = 2 is y = 9x - 13.