What are the points of inflection, if any, of #f(x)= -14x^3 + 19x^2 - x - 2 #?

Question

What are the points of inflection, if any, of #f(x)= -14x^3 + 19x^2 - x - 2     #?

Isabella Ackerman · Accepted Answer

Points of inflection are #(0.64,1.47)# and #(-0.21, -0.812)# To find points of inflection, differentiate the expression to get the gradients, and then find the points where the gradient is zero. #f(x) = -14x^3 +19x^2 - x - 2# #f'(x)=-42x^2 +38x -1# for points of inflection, #-42x^2+38x-1 = 0# #42x^2 -38x +1 = 0# This does not factorise so use the quadratic equation #x = (-b+-sqrt(b^2-4ac))/(2a)# #x=(38+-sqrt(1444-168))/84# #x~~(38+-35.72)/84# #x~~0.64# or #x~~-0.21# Substitute back into the original expression to find the values of #f(x)# Points of inflection are #(0.64,1.47)# and #(-0.21, -0.812)#

Ezekiel Alexander · Answer

undefined To find the points of inflection, we first need to find the second derivative of the function and then solve for where the second derivative equals zero. The second derivative of $ f(x) = -14x^3 + 19x^2 - x - 2 $ is $ f''(x) = -84x + 38 $. Setting $ f''(x) = 0 $ and solving for $ x $, we get:

$$ -84x + 38 = 0 $$
$$ -84x = -38 $$
$$ x = \frac{38}{84} = \frac{19}{42} $$

Thus, the point of inflection of the function $ f(x) $ is $ \left(\frac{19}{42}, f\left(\frac{19}{42}\right)\right) $.