How do you find the maximum, minimum and inflection points and concavity for the function #f(x)=18x^3+5x^2-12x-17#?

Question

Evelyn Agard · Accepted Answer

Below

#f(x)=18x^3+5x^2-12x-17# #f'(x)=54x^2+10x-12# #f''(x)=108x+10# For maximum or minimum points, #f'(x)=0# #54x^2+10x-12=0# #27x^2+5x-6=0# #x=(-5+-sqrt(25+648))/54# #x=(-5+-sqrt673)/54# #x=(-5+sqrt673)/54# or #x=(-5-sqrt673)/54# To determine whether the point is maximum or minimum, At #((-5+sqrt673)/54,-19.85)#, #f''(x)=51.88>0# Therefore, it is minimum and concave up at #((-5+sqrt673)/54,-19.85)# At #((-5-sqrt673)/54,-0.57)# #f''(x)=-51.88 <0# Therefore, it is maximum and concave down at #((-5-sqrt673)/54,-0.57)# For inflexion points, #f''(x)=0# #108x+10=0# #x=-10/108=-5/54# Test #(-5/54,-15.86)# #f''(0)=0+10=10>0# #f''(-5/54)=0# #f''(-1/2)=-44 <0# Therefore, since there is a change in concavity, a point of inflexion occurs at #(-5/54,-15.86)#

Scarlett Allison · Answer

undefined To find the maximum, minimum, inflection points, and concavity for the function $ f(x) = 18x^3 + 5x^2 - 12x - 17 $: 1. **To find critical points:** - Find the first derivative $ f'(x) $. - Set $ f'(x) = 0 $ and solve for $ x $ to find critical points. 2. **To find maximum and minimum points:** - Use the second derivative test on critical points. - Find the second derivative $ f''(x) $. - Substitute each critical point into $ f''(x) $. - If $ f''(x) > 0 $, it's a minimum point. If $ f''(x) < 0 $, it's a maximum point. 3. **To find inflection points:** - Find the second derivative $ f''(x) $. - Set $ f''(x) = 0 $ and solve for $ x $ to find possible inflection points. - Confirm the change in concavity around these points. 4. **To determine concavity:** - Examine the sign of $ f''(x) $. - If $ f''(x) > 0 $, the function is concave up. - If $ f''(x) < 0 $, the function is concave down. After following these steps, you can identify maximum points, minimum points, inflection points, and the concavity of the function.