How do you find the local maximum and minimum values of #f(x)=2x^3 + 5x^2 - 4x - 3#?

Question

Cameron Alexander · Accepted Answer

#f(-2)=9# is local Maxima, #f(1/3)=-100/27# is local

Minima.

We know that, for local Extreme values, #f'(x)=0.# Also, #f''(x)<0# for Maxima, and, #f''(x) >0# for Minima. #f(x)=2x^3+5x^2-4x-3# # rArr f'(x)=6x^2+10x-4, &, f''(x)=12x+10.# # f'(x)=0 rArr 2(3x^2+5x-2)=0.# # rArr 2(x+2)(3x-1)=0.# # rArr x=-2, x=1/3.# Now, #f''(-2)=-24+10=-14 < 0.# #:. f# has a local maxima at #x=-2,# &, it is, #f(-2)=-16+20+8-3=9.# Also, #f''(1/3)=4+10=14 > 0.# #:. f# has a local minima at #x=1/3,# which is, #f(1/3)=2/27+5/9-4/3-3=-100/27.# Thus, #f(-2)=9# is local Maxima, #f(1/3)=-100/27# is local Minima. Enjoy Maths.!

Samantha Adkison · Answer

undefined To find the local maximum and minimum values of $ f(x) = 2x^3 + 5x^2 - 4x - 3 $: 1. Find the critical points by setting the derivative of $ f(x) $ equal to zero and solving for $ x $. $ f'(x) = 6x^2 + 10x - 4 $ Set $ f'(x) = 0 $ and solve for $ x $: $ 6x^2 + 10x - 4 = 0 $ $ 3x^2 + 5x - 2 = 0 $ Use the quadratic formula: $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $ 2. After solving for $ x $, you'll get critical points $ x_1 $ and $ x_2 $. 3. Determine the nature of these critical points by evaluating the second derivative of $ f(x) $. $ f''(x) = 12x + 10 $ 4. Substitute the critical points $ x_1 $ and $ x_2 $ into $ f''(x) $ and determine whether $ f''(x_1) > 0 $ or $ f''(x_2) < 0 $. 5. If $ f''(x_1) > 0 $, then $ x_1 $ corresponds to a local minimum, and if $ f''(x_2) < 0 $, then $ x_2 $ corresponds to a local maximum. 6. Calculate $ f(x_1) $ and $ f(x_2) $ to find the values of the local minimum and maximum, respectively. That's how you find the local maximum and minimum values of the function $ f(x) $.