Question

Find the equation of the tangent line to the graph of g(x)=f′(x) at x=−1 when f(x)=4x³+3x² +2x. (Express numbers in exact form. Use symbolic notation and fractions where needed. Express equation in terms of y and x, where y is the dependent variable and x is the independent variable.)

Answer 1

To find the equation of the tangent line to the graph of g(x) = f'(x) at x = -1, we first find the derivative of f(x). Then, we substitute x = -1 into the derivative to find the slope of the tangent line. Finally, using the point-slope form of the equation of a line, we can write the equation of the tangent line as y - 8 = 8(x + 1).

Step-by-step explanation:

To find the equation of the tangent line to the graph of g(x) = f'(x) at x = -1, we need to find the derivative of f(x) first. The given function is f(x) = 4x^3 + 3x^2 + 2x. Taking the derivative of f(x), we get f'(x) = 12x^2 + 6x + 2.

To find the equation of the tangent line at x = -1, we substitute -1 into f'(x) to find the slope of the tangent line. Therefore, the slope is given by f'(-1) = 12(-1)^2 + 6(-1) + 2 = 12 - 6 + 2 = 8.

Now, we have the slope of the tangent line, which is 8. Using the point-slope form of the equation of a line, we can write the equation of the tangent line as y - y1 = m(x - x1). Substituting the values we know, x1 = -1, y1 = f'(-1) = 8, and m = 8, we can simplify the equation to y - 8 = 8(x + 1).