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

Question

Emilia Aguilar · Accepted Answer

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#.

#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#.

Elijah Abram · Answer

The function is growing.

And now, to complete the set using logs #y = (x+3)(x-3)(3x-1)# Take both sides' logs. #ln(y) = ln((x+3)(x-3)(3x-1))# Make use of log properties #ln(y) = ln(x+3)+ln(x-3)+ln(3x-1)# Distinguish the two 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 role is expanding.

Scarlett Allison · Answer

undefined To determine if the function $ f(x) = (x+3)(x-3)(3x-1) $ is increasing or decreasing at $ x = 2 $, we need to analyze the sign of the derivative $ f'(x) $ at $ x = 2 $. If $ f'(x) > 0 $, the function is increasing at that point. If $ f'(x) < 0 $, the function is decreasing at that point. If $ f'(x) = 0 $, the function has a stationary point.

To find $ f'(x) $, we differentiate $ f(x) $ with respect to $ x $ using the product rule:

$$ f'(x) = (x-3)(3x-1) + (x+3)(3x-1) + (x+3)(x-3)(3) $$

$$ = 3x^2 - x - 3 + 3x^2 - 9x + 3 + 3x^2 - 9 $$

$$ = 9x^2 - 9x - 9 $$

Now, evaluate $ f'(2) $:

$$ f'(2) = 9(2)^2 - 9(2) - 9 = 36 - 18 - 9 = 9 > 0 $$

Since $ f'(2) > 0 $, the function $ f(x) $ is increasing at $ x = 2 $.