What is the solution set of –|–x| = –12?

Mathematics

2 answers:

tatiyna3 years ago

7 0

It’s would be positive x=12

Akimi4 [234]3 years ago

4 0

Answer:

I-xI=12

-x=12

-x=-12

x=-12 or -12.

{-12, 12}

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Maclaurin series is a special form of Taylor series where a = 0,

$f(x)=\sum\limits^\infty_{n=0}\frac{f^{(n)}(0)(x-a)^n}{n!}=\sum\limits^\infty_{n=0}\frac{f^{(n)}(0)x^n}{n!}$

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For $f(x)=xe^{2x}$ ,

$f\left(x\right)=0\cdot \:e^{2\cdot \:0}+\frac{\frac{d}{dx}\left(xe^{2x}\right)\left(0\right)}{1!}x+\frac{\frac{d^2}{dx^2}\left(xe^{2x}\right)\left(0\right)}{2!}x^2+\frac{\frac{d^3}{dx^3}\left(xe^{2x}\right)\left(0\right)}{3!}x^3+\ldots$

Where,

$0\cdot \:e^{2\cdot \:0}\quad :\quad 0\\\\\frac{d}{dx}\left(xe^{2x}\right)\left(0\right)\quad :\quad 1\\\\\frac{d^2}{dx^2}\left(xe^{2x}\right)\left(0\right)\quad :\quad 4\\\\\frac{d^3}{dx^3}\left(xe^{2x}\right)\left(0\right)\quad :\quad 12\\\\\frac{d^4}{dx^4}\left(xe^{2x}\right)\left(0\right)\quad :\quad 32\\\\\frac{d^5}{dx^5}\left(xe^{2x}\right)\left(0\right)\quad :\quad 80$

Thus,

$f\left(x\right)=0+\frac{1}{1!}x+\frac{4}{2!}x^2+\frac{12}{3!}x^3+\frac{32}{4!}x^4+\frac{80}{5!}x^5+\ldots=\\\\=x+2x^2+2x^3+\frac{4}{3}x^4+\frac{2}{3}x^5+\ldots$