What is the point(s) of inflection of the curve of #g(x) = -4x^3-2x^2+x - 1#?

1 Answer
Steve M
Jul 14, 2017

There is a non-stationary point of inflection at #(-1/6, -65/54)#

Explanation:

We have:

# g(x) = -4x^3-2x^2+x-1 #
graph{-4x^3-2x^2+x-1 [-5, 5, -5, 5]}

We would normally look for critical points, that is coordinates where #dy/dx=0#, however the question does not require this so it will be skipped.

The first derivative is then:

# g'(x) = -12x^2-4x+1 #

So the second derivative is then:

# g''(x) = -24x-4 #

We look for inflection points, which are coordinates where the second derivative vanishes:

# g''(x) = 0 => -24x-4 = 0 #
# :. 24x=-4 #
# :. x = -1/6 #

A point of inflection is graded as a stationary point of inflection of the first derivative vanished at the point, otherwise as a non-stationary point of inflection

So when #x=-1/6#, we have:

# g(-1/6) = 4(1/216) - 2(1/36) -1/6 -1 #
# " " = -65/54 #

And:

# g'(-1/6) = -12(1/36) + 4(1/6) + 1 #
# " " = 4/3 #

Hence, There is a non-stationary point of inflection at #(-1/6, -65/54)#

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