Find the value(s) of x such that f''(x)=0
[tex] f_{(x)=} \frac{x}{ x^{2} +2} using \frac{dy}{dx} = \frac{u'v-uv'}{ v^{2} }
[/tex]

[tex]f(x)=\dfrac x{x^2+2}\implies f'(x)=\dfrac{2-x^2}{(x^2+2)^2}\implies f''(x)=\dfrac{2x^3-12x}{(x^2+2)^3}[/tex]

The denominator is always positive, so any [tex]x[/tex] for which [tex]f''(x)=0[/tex] is determined only by the numerator:

[tex]2x^3-12x=0\implies f''(x)=0[/tex]
[tex]2x^3-12x=2x(x^2-6)=2x(x-\sqrt6)(x+\sqrt6)=0\implies x=0,\pm\sqrt6[/tex]

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Аноним Аноним · Answer 2 · 2018-01-12T20:47:18+01:00

25q + 10d + 5n = 127cents (not possible for the sum to be this. (sum needs to end in 0 or 5)"One dollar and eighty-seven cents. That was all. And sixty cents of it was in pennies." Let q be the number of quarters, d be the number of dimes, and n be the number of nickels. A. Create and simplify the equation that represents the value of the coins. Make it an equation that expresses the value of quarters, dimes, and nickels. B. Is it possible to have exactly 60 pennies in $1.87? Explain.for ex25 + 10 + 50r50 + 10 + 5

Find the value(s) of x such that f''(x)=0 [tex] f_{(x)=} \frac{x}{ x^{2} +2} using \frac{dy}{dx} = \frac{u'v-uv'}{ v^{2} } [/tex]

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Find the value(s) of x such that f''(x)=0
[tex] f_{(x)=} \frac{x}{ x^{2} +2} using \frac{dy}{dx} = \frac{u'v-uv'}{ v^{2} }
[/tex]