How do you find the maxima and minima of the function #f(x)=x^3+3x^2-24x+3#?

Question

Emily Ammerman · Accepted Answer

maxima at #(-4,83)#
minima at #(2,-25)#

We have: # f'(x) = 3x^2+6x-24 # At a max/min (turning point) #f'(x)= 0 => 3x^2+6x-24 = 0# # :. \ \ \ \ x^2+2x-8=0 # # :. (x-2)(x+4) = 0# # :. x=-4,2,# When #x=-4=> f(-4)=-64+48+96+3 = 83 # When #x=2 \ \ \ \ \=> f(2) \ \ \ \ \ =8+12-48+3 \ \ \ \ \ \ \ = -25 # To determine the nature of the turning points we look at the second derivative. Differentiating again wrt #x#: # f''(x) = 6x+6 # When #x=-4 => f''(-4) < 0 => # maximum When #x=2 \ \ \ \ \=> f''(2) > 0 \ \ \ \ \ \ => # minimum So the maxima and minima are: maxima at #(-4,83)# minima at #(2,-25)# We can confirm these results graphically: graph{x^3+3x^2-24x+3 [-10, 10, -50, 100.]}

Emily Ammerman · Answer

undefined To find the maxima and minima of the function $ f(x) = x^3 + 3x^2 - 24x + 3 $, you need to follow these steps:

1. Find the derivative of the function, $ f'(x) $.
2. Set the derivative equal to zero and solve for $ x $ to find critical points.
3. Use the second derivative test or the first derivative test to determine whether these critical points are maxima or minima.

Let's go through these steps:

1. Find the derivative of $ f(x) $:
$$ f'(x) = 3x^2 + 6x - 24 $$

2. Set the derivative equal to zero and solve for $ x $:
$$ 3x^2 + 6x - 24 = 0 $$
$$ x^2 + 2x - 8 = 0 $$
$$ (x + 4)(x - 2) = 0 $$

Solving for $ x $:
$$ x + 4 = 0 \implies x = -4 $$
$$ x - 2 = 0 \implies x = 2 $$

So, the critical points are $ x = -4 $ and $ x = 2 $.

3. Use the first derivative test or the second derivative test to determine whether these critical points are maxima or minima.

First Derivative Test:
- For $ x = -4 $: $ f'(-4) = 3(-4)^2 + 6(-4) - 24 = 48 - 24 - 24 = 0 $ (changing from increasing to decreasing), indicating a local maximum.
- For $ x = 2 $: $ f'(2) = 3(2)^2 + 6(2) - 24 = 12 + 12 - 24 = 0 $ (changing from decreasing to increasing), indicating a local minimum.

Therefore, the function has a local maximum at $ x = -4 $ and a local minimum at $ x = 2 $.