What are the points of inflection, if any, of #f(x)=4x^3 + 21x^2 - 294x +7

#?

Question

What are the points of inflection, if any, of #f(x)=4x^3 + 21x^2 - 294x +7

#?

Isaiah Allen · Accepted Answer

#(7/2,-2373/4) and (-7,1722)# are points of minimum and maximum respectively, though none are at a point where inflection occurs, but the graph continues going in the same general direction.

An inflection occurs when #f'(x)=0# To determine this, we must differentiate #f(x)dx# #f(x)=4x^3 + 21x^2 - 294x +7# #f'(x)=12x^2+42x-294# #12x^2+42x-294=0# #6x^2+21x-147=0# Using the quadratic formula: #x=(-b+-sqrt(b^2-4ac))/(2a)# We get: #x=(-21+-sqrt(21^2-4(6*-147)))/(2(6))# #x=(-21+-sqrt(441-4(-882)))/(12)# #x=(-21+-sqrt(441+3528))/(12)# #x=(-21+-sqrt(3969))/(12)# #x=(-21+63)/(12)or(-21-63)/(12)# #x=42/12or-84/12# #x=7/2 or -7# #f(7/2)=4(7/2)^3 + 21(7/2)^2 - 294(7/2)+7=-2373/4# #f(-7)=4(-7)^3 + 21(-7)^2 - 294(-7)+7=1722# The points of inflection are: #(7/2,-2373/4) and (-7,1722)# But this only indicates whether a given point is an inflection, a maximum, or a minimum. The second derivative can help us with this. If #f''(x)=0# it is inflection, #f''(x)>0# is a minimum, #f''(x)<0# is a maximum. #f''(x)=24x+42# #f''(7/2)=24(7/2)+42=126# (Therefore a minimum) #f''(-7)=24(-7)+42=-126# (Therefore a maximum) There are only minimum and maximum points, not actual points of inflection.

Elijah Abram · Answer

undefined To find the points of inflection, you need to calculate the second derivative of the function f(x). Then, set the second derivative equal to zero and solve for x to find the points where the concavity changes. Finally, check the concavity of the function around these points to confirm if they are points of inflection.

First derivative of f(x): f'(x) = 12x^2 + 42x - 294
Second derivative of f(x): f''(x) = 24x + 42

To find the points of inflection:
24x + 42 = 0
x = -42/24
x = -7/4

So, the potential point of inflection is at x = -7/4.

To check if this is a point of inflection, determine the concavity of the function around x = -7/4:
Choose a value of x less than -7/4, say x = -2, and substitute it into the second derivative:
f''(-2) = 24(-2) + 42 = -48 + 42 = -6

Choose a value of x greater than -7/4, say x = 0, and substitute it into the second derivative:
f''(0) = 24(0) + 42 = 42

Since the sign of the second derivative changes from negative to positive at x = -7/4, the point (-7/4, f(-7/4)) is a point of inflection.

Therefore, the point of inflection for the function f(x) = 4x^3 + 21x^2 - 294x + 7 is (-7/4, f(-7/4)).