What are the critical values, if any, of #f(x)= 5x + 6x ln x^2#?

Question

1509 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-05T12:39:28

# x = e^(-17/12) ~~ 0.659241 \ \ (6 \ dp)#

We have:

# f(x) = 5x+6xln(x^2) #

Which, using the properties of logarithms, we can write as:

# f(x) = 5x+6x(2)ln(x) #
# \ \ \ \ \ \ \ = 5x+12xln(x) #

Then, differentiating wrt #x# by applying the product rule we get:

# f'(x) = 5 + (12x)(1/x) + (12)(lnx) #
# \ \ \ \ \ \ \ \ \ = 5+12+12lnx #
# \ \ \ \ \ \ \ \ \ = 17+12lnx #

At a critical point, we requite that the first derivative vanishes, thus we require that:

# f'(x) = 0 #

# :. 17+12lnx = 0 #
# :. lnx = -17/12 #
# :. x = e^(-17/12) ~~ 0.659241 \ \ (6 \ dp)#