f(x)=(x^2+1)(1-x-x^3).Determine the equation of the line tangent to f(x) at the point (1,-2).
To determine the equation of the line tangent to the function f(x) at the point (1,-2), we will need to find the slope of the tangent line and the point of tangency.
Step 1: Find the derivative of f(x).
Start by finding the derivative of f(x) using the product rule and the power rule of differentiation. Let's do that:
f(x) = (x^2 + 1)(1 - x - x^3)
Recall that the product rule states that the derivative of f(x)g(x) is f'(x)g(x) + f(x)g'(x), where f'(x) is the derivative of f(x) and g'(x) is the derivative of g(x). Apply the product rule:
f'(x) = [2x(1 - x - x^3) + (x^2 + 1)(-1 - 1 - 3x^2)]
= [2x - 2x^2 - 2x^3 - x^2 - 1 - 3x^2]
= [-3x^3 - 4x^2 + 2x - 1]
Step 2: Find the slope of the tangent line.
To find the slope of the tangent line, substitute x = 1 into the derivative f'(x) we found in Step 1:
f'(1) = [-3(1)^3 - 4(1)^2 + 2(1) - 1]
= [-3 - 4 + 2 - 1]
= [-6]
Therefore, the slope of the tangent line is -6.
Step 3: Find the y-coordinate of the tangent point.
To find the y-coordinate of the tangent point, substitute x = 1 into the original function f(x):
f(1) = (1^2 + 1)(1 - 1 - 1^3)
= (1 + 1)(1 - 1 - 1)
= (2)(-1)
= -2
Therefore, the y-coordinate of the tangent point is -2.
Step 4: Write the equation of the tangent line.
Using the point-slope form of a line, we can write the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values we found:
y - (-2) = -6(x - 1)
y + 2 = -6x + 6
y = -6x + 4
Therefore, the equation of the line tangent to f(x) at the point (1,-2) is y = -6x + 4.