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

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Answer 1 · 2017-01-30T23:02:11

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.]}