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Colt1911 [192]
1 year ago
14

\ \textgreater \ 0" alt="5x^{2}+|x+1|\ \textgreater \ 0" align="absmiddle" class="latex-formula">
Can someone explain me how this is done? My book says that the answer is R (all numbers), but i get [-1; infinity) and two other roots.
Mathematics
1 answer:
Pavel [41]1 year ago
7 0
5x² + |x+1| > 0
5x² + x + 1 > 0
x+1 > 0 because it came from absolute
5x² > 0 because it’s squared

When you try a number you should put it in the original equality
5x² + |x+1|
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What are the vertical and horizontal asymptotes for the function f(x)=<br> 3x2/x2-4
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Answer:  f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.

Step-by-step explanation:

Given function: f(x)=\dfrac{3x^2}{x^2-4}

The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.

i.e. x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2

Hence, f(x) will have vertical asymptotes at x=-2 and x=2.

To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.

So, the graph will horizontal asymptote at y=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}

i.e. y=\dfrac{3}{1}=3

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3 0
3 years ago
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3 years ago
I need help with this calculus question
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