Find the points of inflection. f(x) = 2x^{3} + 9x^{2} + 12x A) (0,0) B) (-3,108) C)...


Find the points of inflection.

{eq}\displaystyle\;f(x) = 2x^{3} + 9x^{2} + 12x {/eq}

A) {eq}\displaystyle\;\left(0,0\right) {/eq}

B) {eq}\displaystyle\;\left(-3,108\right) {/eq}

C) {eq}\;\left(-\frac{3}{2}, -\frac{9}{2}\right) {/eq}

D) {eq}\textrm{ No points of inflection exist.} {/eq}

Inflection Points:

For some functions, there are points that we call "inflection points". These points are values where the function's shape changes concavity. The points of inflection for any function can be determined by determining the points where its second derivative is equal to zero.

Answer and Explanation:

Given the function:

{eq}\displaystyle \rm f(x) = 2x^3 + 9x^2 + 12x {/eq}

Let us determine the second derivative of this function. We recall that:

{eq}\displaystyle \rm \frac{d}{dx} \Big[x^n \Big] = nx^{n-1} {/eq}

The first derivative is thus:

{eq}\displaystyle \rm \frac{df}{dx} = 6x^2 + 18x + 12 {/eq}

The second derivative is:

{eq}\displaystyle \rm \frac{d^2 f}{dx^2} = 12x + 18 {/eq}

The points of inflection can be determined if we set:

{eq}\displaystyle \rm \frac{d^2 f}{dx^2} = 0 {/eq}

We set our line to zero:

{eq}\displaystyle \rm 0 = 12x + 18 {/eq}

{eq}\displaystyle \rm 12x = -18 {/eq}

{eq}\displaystyle \rm x = -\frac{18}{12} {/eq}

We get:

{eq}\displaystyle \rm x = -\frac{3}{2} {/eq}

Since there is only one x value here that matches the choices, are answer should thus be:

{eq}\displaystyle \rm \boxed{\rm (x,y) = \left(-\frac{3}{2}, -\frac{9}{2} \right)} {/eq}

Learn more about this topic:

Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5

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