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To find the inflection points of the function [tex]\( f(x)=\frac{x^2}{7 x^2+8} \)[/tex], we need to analyze the second derivative of this function. Inflection points occur where the second derivative changes sign, which corresponds to the points where the second derivative is equal to zero.
The process is as follows:
1. First Derivative: Calculate the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{x^2}{7x^2 + 8} \][/tex]
Using the quotient rule:
[tex]\[ f'(x) = \frac{(2x)(7x^2 + 8) - (x^2)(14x)}{(7x^2 + 8)^2} = \frac{14x^3 + 16x - 14x^3}{(7x^2 + 8)^2} = \frac{16x}{(7x^2 + 8)^2} \][/tex]
2. Second Derivative: Compute the second derivative from the first derivative:
[tex]\[ f''(x) = \frac{d}{dx} \left( \frac{16x}{(7x^2 + 8)^2} \right) \][/tex]
Applying the quotient rule again:
[tex]\[ f''(x) = \frac{(16)(7x^2 + 8)^2 - (16x)(2)(7x^2 + 8)(14x)}{(7x^2 + 8)^4} = \frac{16(7x^2 + 8)^2 - 448x^2(7x^2 + 8)}{(7x^2 + 8)^4} \][/tex]
Simplify the expression in the numerator:
[tex]\[ 16(7x^4 + 16x^2 + 8) - 896x^4 = 112x^4 + 128x - 896x^4 = 128x - 784x^4 \][/tex]
So the second derivative can be represented as:
[tex]\[ f''(x) = \frac{16(7x^4 + 16x^2 - 56x^4)}{(7x^2 + 8)^2} = \frac{16(-49x^4 + 16x^2)}{(7x^2 + 8)^3} = \frac{16(-49x^4 + 16x^2)}{(7x^2 + 8)^3} \][/tex]
3. Setting the Second Derivative Equal to Zero: Solve for [tex]\( x \)[/tex] in:
[tex]\[ -784x^2(7x^2 - 8) = 0 \][/tex]
This equation simplifies to:
[tex]\[ 112x^4 + 128x^2 = 0 \][/tex]
Since [tex]\(-49x^2(7x^2 - 8) = 0\)[/tex]
[tex]\[ 784x^4 - 112(7x^2 - 8)=0 784x^4 + 128x^2 \][/tex]
This gives roots at:
[tex]\[ x = \pm \frac{2\sqrt{42}}{21} \][/tex]
Therefore, the [tex]\( x \)[/tex] values of the inflection points of [tex]\( f(x) = \frac{x^2}{7x^2 + 8} \)[/tex] are:
[tex]\[ x = -\frac{2\sqrt{42}}{21} \text{ and } x = \frac{2\sqrt{42}}{21} \][/tex]
The process is as follows:
1. First Derivative: Calculate the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{x^2}{7x^2 + 8} \][/tex]
Using the quotient rule:
[tex]\[ f'(x) = \frac{(2x)(7x^2 + 8) - (x^2)(14x)}{(7x^2 + 8)^2} = \frac{14x^3 + 16x - 14x^3}{(7x^2 + 8)^2} = \frac{16x}{(7x^2 + 8)^2} \][/tex]
2. Second Derivative: Compute the second derivative from the first derivative:
[tex]\[ f''(x) = \frac{d}{dx} \left( \frac{16x}{(7x^2 + 8)^2} \right) \][/tex]
Applying the quotient rule again:
[tex]\[ f''(x) = \frac{(16)(7x^2 + 8)^2 - (16x)(2)(7x^2 + 8)(14x)}{(7x^2 + 8)^4} = \frac{16(7x^2 + 8)^2 - 448x^2(7x^2 + 8)}{(7x^2 + 8)^4} \][/tex]
Simplify the expression in the numerator:
[tex]\[ 16(7x^4 + 16x^2 + 8) - 896x^4 = 112x^4 + 128x - 896x^4 = 128x - 784x^4 \][/tex]
So the second derivative can be represented as:
[tex]\[ f''(x) = \frac{16(7x^4 + 16x^2 - 56x^4)}{(7x^2 + 8)^2} = \frac{16(-49x^4 + 16x^2)}{(7x^2 + 8)^3} = \frac{16(-49x^4 + 16x^2)}{(7x^2 + 8)^3} \][/tex]
3. Setting the Second Derivative Equal to Zero: Solve for [tex]\( x \)[/tex] in:
[tex]\[ -784x^2(7x^2 - 8) = 0 \][/tex]
This equation simplifies to:
[tex]\[ 112x^4 + 128x^2 = 0 \][/tex]
Since [tex]\(-49x^2(7x^2 - 8) = 0\)[/tex]
[tex]\[ 784x^4 - 112(7x^2 - 8)=0 784x^4 + 128x^2 \][/tex]
This gives roots at:
[tex]\[ x = \pm \frac{2\sqrt{42}}{21} \][/tex]
Therefore, the [tex]\( x \)[/tex] values of the inflection points of [tex]\( f(x) = \frac{x^2}{7x^2 + 8} \)[/tex] are:
[tex]\[ x = -\frac{2\sqrt{42}}{21} \text{ and } x = \frac{2\sqrt{42}}{21} \][/tex]
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