How do use the first derivative test to determine the local extrema #f(x)=x^3 - 9x^2 + 27x#?

Question

Emilia Aguilar · Accepted Answer

This function has no local extrema.

#f(x) = x^3-9x^2+27x# Has derivative: #f'(x) = 3x^2 -18x+27 = 3(x^2-6x+9) = 3(x-3)^2# Every local extremum occurs at a critical number. (A number, #c#, in the domain of #f# at which either #f'(c) = 0# or #f'(c)# does not exist). The only critical number is #3#, and the derivative does not change sign at #3# (this derivative is always non-negative). So the function has no local extrema.

Elijah Abram · Answer

undefined To use the first derivative test to determine the local extrema of $ f(x) = x^3 - 9x^2 + 27x $, follow these steps:

1. Find the first derivative of the function $ f'(x) $.
$$ f'(x) = 3x^2 - 18x + 27 $$

2. Set the first derivative equal to zero and solve for $ x $ to find the critical points.
$$ 3x^2 - 18x + 27 = 0 $$
$$ x^2 - 6x + 9 = 0 $$
$$ (x - 3)^2 = 0 $$
$$ x = 3 $$

3. Determine the sign of the derivative in intervals around the critical point $ x = 3 $ by testing a value in each interval into the first derivative.
   - For $ x < 3 $, choose $ x = 0 $: $ f'(0) = 27 > 0 $, so the function is increasing.
   - For $ x > 3 $, choose $ x = 4 $: $ f'(4) = 27 > 0 $, so the function is increasing.

4. Analyze the results:
   - Before $ x = 3 $, the function is increasing.
   - After $ x = 3 $, the function is increasing.

Since the function is increasing before and after the critical point $ x = 3 $, there are no local extrema.