What are the local extrema, if any, of #f(x) =2x^3 -3x^2+7x-2 #?

Question

Charles Anctil · Accepted Answer

Are no local extremas in #RR^n# for #f(x)# We'll first need to take the derivative of #f(x)#. #dy/dx=2d/dx[x^3]-3d/dx[x^2]+7d/dx[x]-0# #=6x^2-6x+7# So, #f'(x)=6x^2-6x+7# To solve for the local extremas, we must set the derivative to #0# #6x^2-6x+7=0# #x=(6+-sqrt(6^2-168))/12# Now, we've hit a problem. It's that #x inCC# so the local extremas are complex. This is what happens when we start off in cubic expressions, it's that complex zeros can happen in the first derivative test. In this case, there are no local extremas in #RR^n# for #f(x)#.

Scarlett Allison · Answer

undefined To find the local extrema of $ f(x) = 2x^3 - 3x^2 + 7x - 2 $, we first find the derivative of the function and then solve for critical points by setting the derivative equal to zero and solving for $ x $. Afterward, we determine whether these points correspond to local maxima or minima by examining the sign of the second derivative at those points.

1. Find the derivative of $ f(x) $:
$$ f'(x) = 6x^2 - 6x + 7 $$

2. Set the derivative equal to zero and solve for $ x $:
$$ 6x^2 - 6x + 7 = 0 $$

This equation doesn't have real solutions, so there are no critical points.

3. Since there are no critical points, there are no local extrema for the function $ f(x) = 2x^3 - 3x^2 + 7x - 2 $.