For #f(x) =(x^2 + 3x - 10)^3#, what is the equation of the line tangent to #x =5 # and at what point is the tangent line horizontal?

Question

Aurora Alvarado · Accepted Answer

The tangent line will be horizontal at #x = -3/2, -5 and 2#.

This is essentially two problems in one. I will answer where the tangent line is horizontal. We start by differentiating #f(x)#, using the chain rule. Letting #y= u^3# and #u = x^2 + 3x - 10#. Then dy/(du) = 3u^2# and #(du)/dx = 2x + 3# Then: #dy/dx = dy/(du) xx (du)/dx# #dy/dx = 3u^2 xx 2x + 3# #dy/dx = (6x + 9)(x^2 + 3x - 10)^2# The tangent will be horizontal if its slope is #0#. The derivative represents the rate of change of the function at any given point in its domain. We set the derivative to #0# and solve. #0 = (6x + 9)(x^2 + 3x - 10)^2# #0 = 6x + 9 and 0 = x^2 + 3x -10# #x = -9/6, 2, -5 = -3/2, 2, -5# Hopefully this helps!

Elijah Abram · Answer

undefined The equation of the line tangent to f(x) at x = 5 is y = 600x - 7250. The tangent line is horizontal at the point (5, -7250).