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3 years ago

F(x)=X over x^3-2x^2+5x why will this have no zeros

Mathematics

1 answer:

Mumz [18]3 years ago

5 0

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

$\dfrac{x}{x^3-2x^2+5x}=0\iff x=0$

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$

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2 years ago

Can someone help with this please?

ipn [44]

$7^2=7\cdot7=49\\\\8^2=8\cdot8=64\\\\\sqrt{49}$

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Step-by-step explanation:

We want to find $\lim_{x \to 2^-} \frac{|x-2|}{x-2}$ .

By definition:

$|x-2|=\left \{ {{x-2,\:if\:x\:>\:2} \atop {-(x-2),\:if\:x\:$

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$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} \frac{-(x-2)}{x-2}$ .

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3 years ago

F(x)=X over x^3-2x^2+5x why will this have no zeros​

F(x)=X over x^3-2x^2+5x why will this have no zeros