What are the points of inflection, if any, of #f(x)= 9x^3- 5 x^2-2 #?

Question

What are the points of inflection, if any, of #f(x)= 9x^3-  5  x^2-2   #?

Isaiah Allen · Accepted Answer

when #x=5/27#

When a function's second derivative changes signs, from positive to negative or vice versa, it is said to have reached a point of inflection.

Find #f''(x)#:

#f(x)=9x^3-5x^2-2# #f'(x)=27x^2-10x# #f''(x)=54x-10#

The second derivative could from positive to negative or negative to positive when #f''(x)=0#. Find those points:

#54x-10=0#

#x=10/54=5/27#

At this point, the second derivative is only a potential point of inflection; make sure the second derivative truly changes sign.

When #x<5/27#, we can test #f''(0)#:

#f''(0)=-10" ... "<0#

When #x>5/27#, we can test #f''(1)#:

#f''(1)=44" ... ">0#

Since the sign of the second derivative does change around #x=5/27#, it is a point of inflection.

Graphically, we can verify this: the concavity ought to change.

plot{(9x^3-5x^2-2) [-10, 10, -7.2, 2.8]}

Evelyn Agard · Answer

undefined To find the points of inflection of $ f(x) = 9x^3 - 5x^2 - 2 $, we need to find its second derivative and then solve for the values of $ x $ where the second derivative equals zero or is undefined.

$ f'(x) = 27x^2 - 10x $

$ f''(x) = 54x - 10 $

To find the points of inflection, set the second derivative equal to zero and solve for $ x $:

$ 54x - 10 = 0 $

$ x = \frac{10}{54} = \frac{5}{27} $

Therefore, the point of inflection is at $ x = \frac{5}{27} $.