What is the equation of the tangent line of #f(x)=(x^3-3x +1)(x+2) # at #x=1#?

Question

William Abdalla · Accepted Answer

The equation of the tangent line of f at x = 1 is y = -x - 2.

The general equation of the tangent line of a given function #f# at #a in I# where #I# is the part of #RR# where the function is defined is given by the formula : #y = f'(a)(x - a) + f(a)#. We're going to use it here. You first need #f'#, which will instantly give you the coefficient of the variable. #f'(x) = 3(x^2 - 1)(x + 2) - x^3 - 3x + 1# You calculate #f'(1)# which equals to #-1#. You use the formula given above and you will find that #y = -x - 1# is the tangent you were looking for.

Christopher Allers · Answer

undefined To find the equation of the tangent line of f(x)=(x^3-3x +1)(x+2) at x=1, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the product rule: f'(x) = (3x^2 - 3)(x+2) + (x^3 - 3x + 1)(1).

Next, we substitute x=1 into f'(x) to find the slope of the tangent line: f'(1) = (3(1)^2 - 3)(1+2) + (1^3 - 3(1) + 1)(1).

Simplifying, we get f'(1) = 3(3) + (-1) = 8.

So, the slope of the tangent line is 8.

To find a point on the line, we substitute x=1 into f(x): f(1) = (1^3 - 3(1) + 1)(1+2) = 0.

Therefore, a point on the tangent line is (1, 0).

Using the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we can write the equation of the tangent line as: y - 0 = 8(x - 1).

Simplifying, we get y = 8x - 8.

Thus, the equation of the tangent line of f(x)=(x^3-3x +1)(x+2) at x=1 is y = 8x - 8.