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Find the x-value at which the function has a point of inflection. y= \frac{1}{3}x^3 - 4x^2 + 12x + 6

Question:

Find the x-value at which the function has a point of inflection.

{eq}y= \frac{1}{3}x^3 - 4x^2 + 12x + 6 {/eq}

Inflection Points:

We know that at inflection point, the second order derivative is zero. But this point may not be inflection because at this point the curve may not be change. So, we have to check the concavity to the left and right side of this inflection point. If this point is not inflection then it may be local maximum or local minimum.

The following derivative rules are used to solve this problem:

1.) $$\frac{d}{dx} (x^n) = nx^{(n - 1)} $$

2.) $$\frac{d}{dx} (c) = 0 \, \text{ Where c is a constant } $$

Answer and Explanation:

The function is given by:

{eq}y = f(x) = \frac{1}{3}x^3 - 4x^2 + 12x + 6 {/eq}

First order derivative of {eq}f(x) {/eq} with respect to {eq}x {/eq} is:

{eq}\begin{align*} f^{'} (x) &= \frac{1}{3} \frac{d}{dx} (x^3) - 4 \frac{d}{dx} (x^2) + 12 \frac{d}{dx} (x) + \frac{d}{dx} (6) \\ &= \frac{1}{3}(3x^2) - 4(2x) + 12(1) + 0 \hspace{1 cm} \left[ \text{ By using [1]and [2] } \right] \\ &= x^2 - 8x + 12 \end{align*} {/eq}

Second order derivative of {eq}f(x) {/eq} with respect to {eq}x {/eq} is:

{eq}\begin{align*} f^{''} (x) &= \frac{d}{dx} (x^2) - 8 \frac{d}{dx} (x) + \frac{d}{dx} (12) \\ &= 2x - 8(1 ) + 0 \hspace{1 cm} \left[ \text{ By sing [1] and [2] } \right] \\ &= 2x - 8 \end{align*} {/eq}


Now, for finding inflection points, we have to set {eq}f^{''} (x) = 0 {/eq}

Hence:

{eq}f^{''} (x) = 0 \\ \Rightarrow 2x - 8 = 0 \\ \Rightarrow 2x = 8 \\ \Rightarrow x = \frac{8}{2} \\ \Rightarrow x = 4 {/eq}

Now, we take two points which are {eq}x =2 \,\, and \,\, x= 5 {/eq} to the left side and right side of {eq}x = 4 {/eq} respectively.

At, {eq}x = 2 {/eq}

{eq}f^{''} (x) = 2(2) - 8 = -4 {/eq}

And:

At, {eq}x = 5 {/eq}

{eq}f^{''} (x) = 2(5) - 8 = 2 {/eq}

From the difference values of {eq}f^{''} (x) {/eq} , we can say that the curve is concave down to the left and concave up to the right of {eq}x=4 {/eq}

So, we can conclude that the concavity of the curve is changes at {eq}x = 4 {/eq} and it is the {eq}x- {/eq} value of inflection point.

Hence:

The {eq}x {/eq}-value at which the function has point of inflection is

{eq}x = 4 {/eq}


Learn more about this topic:

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Understanding Concavity and Inflection Points with Differentiation

from Math 104: Calculus

Chapter 10 / Lesson 6
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