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Find the zeros of the function f(x)=x^2-1/x+4
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Is your function \(x^2 - \frac{1}{x}+4\) or \(\large \frac{ x^2-1}{ x+4}\)
f(x)=(x^2-1)/(x+4)
Second option you posted.
Ok, well first, set it equal to zero.
You have \( \huge \frac{x^2-1}{x+4}=0\)
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next, multiply both sides by \(\large x+4\)
You will be left with \(\large x^2-1=0\) which is simply the difference \(\large (x+1)(x-1) = 0\) now you will have two solutions to solve which you can easily do yoruself: \(\large (x+1) = 0\) and \(\large (x-1)=0\) and your zero's are: x =
The zeros of a function are the x coordinates of the x intercepts of the graph of f.
OK, So x=-1,1?
Yes.
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Oh, You are awesome. Thank you!! :)
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