What are the extrema of #f(x)=2x^3 + 5x^2 - 4x - 3 #?

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Answer 1 · 2016-12-10T12:37:54

#x_1=-2# is a maximum

#x_2=1/3# is a minimum.

First we identify the critical points by equating the first derivative to zero:

#f'(x) = 6x^2 +10x -4 = 0#

giving us:

#x= frac (-5 +- sqrt (25+24)) 6 = (-5 +- 7)/6#

#x_1= -2# and #x_2=1/3#

Now we study the sign of the second derivative around the critical points:

#f''(x) = 12x+10#

so that:

#f''(-2) < 0 # that is #x_1=-2# is a maximum

#f''(1/3) > 0 # that is #x_2=1/3# is a minimum.

graph{2x^3+5x^2-4x-3 [-10, 10, -10, 10]}