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

Question

Charles Anctil · Accepted Answer

#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]}

Christopher Agresta · Answer

undefined To find the extrema of the function $ f(x) = 2x^3 + 5x^2 - 4x - 3 $, we first need to find its critical points. We find the derivative of the function, set it equal to zero, and solve for $ x $ to find critical points. Then, we use the second derivative test to determine whether these critical points correspond to relative minima, relative maxima, or points of inflection.

The derivative of $ f(x) $ is $ f'(x) = 6x^2 + 10x - 4 $. Setting this equal to zero, we get $ 6x^2 + 10x - 4 = 0 $.

Solving this quadratic equation, we find the critical points by applying the quadratic formula:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Substituting $ a = 6 $, $ b = 10 $, and $ c = -4 $, we get:

$$ x = \frac{{-10 \pm \sqrt{{10^2 - 4 \cdot 6 \cdot (-4)}}}}{{2 \cdot 6}} $$

$$ x = \frac{{-10 \pm \sqrt{{100 + 96}}}}{{12}} $$

$$ x = \frac{{-10 \pm \sqrt{{196}}}}{{12}} $$

$$ x = \frac{{-10 \pm 14}}{{12}} $$

$$ x_1 = \frac{4}{12} = \frac{1}{3} $$

$$ x_2 = \frac{-24}{12} = -2 $$

Next, we evaluate the second derivative $ f''(x) = 12x + 10 $.

For $ x = \frac{1}{3} $, $ f''\left(\frac{1}{3}\right) = 12 \cdot \frac{1}{3} + 10 = 14 $, which is positive. Hence, $ x = \frac{1}{3} $ corresponds to a relative minimum.

For $ x = -2 $, $ f''(-2) = 12 \cdot (-2) + 10 = -14 $, which is negative. Hence, $ x = -2 $ corresponds to a relative maximum.

Therefore, the relative minimum occurs at $ x = \frac{1}{3} $ and the relative maximum occurs at $ x = -2 $.

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