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?

1 Answer
Dec 6, 2016

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

Explanation:

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!