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

Question

Isaiah Allen · Accepted Answer

The point of inflection is #=(-0.778,-5.156)##

The first and second derivatives are computed. #f(x)=-3x^3-7x^2+3x# #f'(x)=-9x^2-14x+3# #f''(x)=-18x-14# When is the turning point? #f''(x)=0# #-18x-14=0#, #=>#, #x=-14/18=-7/9# As a result, the turning point is #(-7/9, f(-7/9))=(-0.778,-5.156)# We could create a chart. #color(white)(aaaa)##x##color(white)(aaaa)##(-oo, -7/9)##color(white)(aaaa)##(-7/9,+oo)# #color(white)(aaaa)##f''(x)##color(white)(aaaaaa)##+##color(white)(aaaaaaaaaaaaa)##-# #color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##uu##color(white)(aaaaaaaaaaaaa)##nn# graph{(y-(-3x^3-7x^2+3x))((x+0.778)^2+(y+5.156)^2-0.01)=0 [-11.24, 6.54, -8.084, 0.805]}

Charles Anctil · Answer

undefined To find the points of inflection, we need to find where the second derivative changes sign. The second derivative of f(x) = -3x^3 - 7x^2 + 3x is f''(x) = -18x - 14. Setting f''(x) equal to zero gives us x = -7/18. This is the only critical point. To determine if it's a point of inflection, we need to check the concavity around it. 
Calculate the second derivative at a point greater than -7/18, for instance, x = 0, we get f''(0) = -14. This indicates the function is concave down to the left of x = -7/18.
Similarly, calculating the second derivative at a point less than -7/18, for instance, x = -1, we get f''(-1) = 4. This indicates the function is concave up to the right of x = -7/18.
Therefore, x = -7/18 is a point of inflection.