Find points of inflection for f(x) = \ln(x^2 + 4).

Question:

Find points of inflection for {eq}f(x) = \ln(x^2 + 4). {/eq}

Points of Inflection:

We comprehend that curvature implies the measurement of the instantaneous rate of change of the path of a point on a curve.

Now:

This curvature alters its path on a curve of a function at inflection points. At these points, we moreover remember that the second order derivative is zero.

Chain Rule of Derivative:

{eq}F^{'} (x) = (f(g(x)))^{'} g^{'} (x) \hspace{1 cm} \text{ Where F(x) = f(g(x))} {/eq}

Quotient Rule of Derivative:

{eq}\frac{d}{dx} (\frac{f(x)}{g(x)}) = \frac{ g(x) \frac{d}{dx} (f(x)) - f(x) \frac{d}{dx} (g(x))}{(g(x))^2} {/eq}

Answer and Explanation:

The function is given by:

{eq}y = f(x) = \ln (x^2 + 4) \hspace{1 cm} \text{(Equation 1)} {/eq}


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

{eq}\begin{align*} f^{'} (x) &= \frac{d}{dx} \left( \ln (x^2 + 4) \right) \\ &= \frac{1}{x^2 + 4} \frac{d}{dx} (x^2 + 4) \hspace{1 cm} \left[ \text{ By using Chain Rule of Derivative and } \frac{d}{dx} (\ln (x)) = \frac{1}{x} \right] \\ &= \frac{1}{x^2 + 4} \left( \frac{d}{dx} (x^2) + \frac{d}{dx} (4) \right) \hspace{1 cm} \left[ \because \frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} (f(x)) + \frac{d}{dx} (g(x)) \right] \\ &= \frac{1}{x^2 + 4} \left( 2x + 0 \right) \hspace{1 cm} \left[ \because \frac{d}{dx} (x^n) = nx^{n-1} \, and \, \frac{d}{dx} (c) = 0 \, \text{ (Where c is a constant)} \right] \\ &= \frac{2x}{x^2 + 4} \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} \left( \frac{2x}{x^2 + 4} \right) \\ &= \frac{(x^2 + 4) \frac{d}{dx} (2x) - 2x \frac{d}{dx} (x^2 + 4)}{(x^2 + 4)^2} \hspace{1 cm} \left[ \text{ By using Quotient Rule of Derivative} \right] \\ &= \frac{(2x^2 + 8) \frac{d}{dx} (x) - 2x \left( \frac{d}{dx} (x^2) + \frac{d}{dx} ( 4) \right)}{(x^2 + 4)^2} \hspace{1 cm} \left[ \because \frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} (f(x)) + \frac{d}{dx} (g(x)) \, and \, \frac{d}{dx} (cf(x)) = c \frac{d}{dx} (f(x)) \, \text{( Where c is a constant and f(x) and g(x) both are functions)} \right] \\ &= \frac{(2x^2 + 8)(1) - 2x \left( 2x + 0 \right)}{(x^2 + 4)^2} \hspace{1 cm} \left[ \because \frac{d}{dx} (x^n) = nx^{n-1} \, and \, \frac{d}{dx} (c) = 0 \, \text{ (Where c is a constant)} \right] \\ &= \frac{2x^2 + 8 - 4x^2}{(x^2+4)^2} \\ &= \frac{8 - 2x^2}{(x^2 + 4)^2} \end{align*} {/eq}


For finding points of inflection, we have to set {eq}f^{"} (x) = 0 {/eq}

Hence:

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


Inserting {eq}x = 2 {/eq} into the (Equation 1), we have:

{eq}y = \ln (2^2 + 4) = \ln (8) {/eq}

Again:

Inserting {eq}x = -2 {/eq} into the (Equation 1), we have:

{eq}y = \ln ((-2)^2 + 4) = \ln (8) {/eq}


Hence:

The points of inflection for the given function are {eq}\left( 2 , \ln (8) \right) \, and \, \left( -2 , \ln (8) \right) {/eq}


Learn more about this topic:

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Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
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