How do you find the equation of the tangent line to a derivative of a function #3x^2-12# at f'(1)?

Question

Christopher Agresta · Accepted Answer

You're halfway there. The Linear Approximation Method is:
#f_T(x) = f(a) + f'(a)(x-a)# #= f(1) + f'(1)(x-1)# #= (3x^2 - 12) + (6x)(x-1)# 
at #x = 1# (except for the #x# on the far right) #= (3(1)^2 - 12) + (6(1))(x-1)# #= -9 + 6(x-1)# #= -9 + 6x- 6# #f_T(x) = 6x - 15# Here are the graphs.

William Abdalla · Answer

undefined To find the equation of the tangent line to the derivative of a function at a specific point, you can follow these steps:

1. Find the derivative of the original function.
2. Evaluate the derivative at the given point to find the slope of the tangent line.
3. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, to write the equation of the tangent line.

In this case, the original function is f(x) = 3x^2 - 12.

1. Find the derivative of f(x) using the power rule: f'(x) = 6x.
2. Evaluate the derivative at x = 1: f'(1) = 6(1) = 6. This gives us the slope of the tangent line.
3. Use the point-slope form with the given point (1, f'(1)) = (1, 6): y - 6 = 6(x - 1).

Therefore, the equation of the tangent line to the derivative of the function 3x^2 - 12 at f'(1) is y - 6 = 6(x - 1).