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avanturin [10]
2 years ago
12

%2B2%7D%20%29%5E%7B%5Cfrac%7Bx%7D%7B2%7D%20%7D" id="TexFormula1" title="\lim_{x \to +\infty} (\frac{x^{2}-3x-1 }{x^{2} -4x+2} )^{\frac{x}{2} }" alt="\lim_{x \to +\infty} (\frac{x^{2}-3x-1 }{x^{2} -4x+2} )^{\frac{x}{2} }" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
Natalka [10]2 years ago
7 0

Step-by-step explanation:

I know but I m feeling bore

Effectus [21]2 years ago
6 0
I solved on my phone it was a easy question just used a calculator to write up the fraction and infinite etc

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B. First off , standard form of a 2nd degree equation is Ax^2 + Bx + C. So look at the coefficient of Ax^2 which is -2.
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A. To find the vertex (in this case maximum),
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3 years ago
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soldier1979 [14.2K]

Use\\\\\left(\dfrac{h(x)}{g(x)}\right)'=\dfrac{h'(x)g(x)-h(x)g'(x)}{[g(x)]^2}\\\\\text{We have}\\\\h(x)=x^3-5x\to h'(x)=3x^2-5\\\\g(x)=x^2-1\to g'(x)=2x\\\\f(x)=\dfrac{x^3-5x}{x^2-1}\to f'(x)=\left(\dfrac{x^3-5x}{x^2-1}\right)'\\\\=\dfrac{(x^3-5x)'(x^2-1)-(x^3-5x)(x^2-1)'}{(x^2-1)^2}\\\\=\dfrac{(3x^2-5)(x^2-1)-(x^3-5x)(2x)}{(3x^2-5)^2}\\\\=\dfrac{3x^4-3x^2-5x^2+5-2x^4+10x^2}{(x^2-1)^2}\\\\=\dfrac{x^4+2x^2+5}{(x^2-1)^2}\\\\Answer:\ \left(\dfrac{x^3-5x}{x^2-1}\right)'=\dfrac{x^4+2x^2+5}{(x^2-1)^2}

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Answer:

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