Is #f(x)=(x+3)(x-3)(3x-1)# increasing or decreasing at #x=2#?

2 Answers
Nov 21, 2015

Like all functions it is constant at a value in its domain. Its derivative is continuous and positive at #2#, so it is increasing in some open interval containing #2#.

Explanation:

#f(x)=(x^2-9)(3x-1)#, so #f(2) = 25#.

#f'(x) = 2x(3x-1)+3(x^2-9)#

# = 9x^2-2x-27#

#f'(2) = 5#, and #f'(x)# is continuous, so #f'(x)# is positive for all #x# in some open interval containing #2#.

Therefore, #f(x)# if increasing in some open interval containing #2#.

Nov 21, 2015

The function is growing.

Explanation:

And now, using logs, to round out the set

#y = (x+3)(x-3)(3x-1)#

Take the log of both sides

#ln(y) = ln((x+3)(x-3)(3x-1))#

Use log properties

#ln(y) = ln(x+3)+ln(x-3)+ln(3x-1)#

Differentiate both sides

#y^'/y = 1/(x+3) + 1/(x-3) + 3/(3x-1)#

Multiply both sides by #y#

#y^' = (x-3)(3x-1) + (x+3)(3x-1) + 3(x+3)(x-3)#

Put in, #x = 2# and see the value

#y^' = (2-3)(6-1) + (2+5)(6-1) + 3(2+3)(2-3)#
#y^' = -5 + 7*5 - 3*5 = -5 - 15 + 35 = 15#

The function is growing.