Ok, so here's my actual problem. . .
\(\displaystyle \L{}g(x) = \frac{-2x^2 + x}{x}\)
From the graph of this function, I was asked to find three things.
1) \(\displaystyle \L{}\lim_{x \to 0} g(x)\)
2) \(\displaystyle \L{}\lim_{x \to -1} g(x)\)
3) A simpler function that agrees with the given function at all but one point.
So, I looked at the graph and found this:
1) 1
2) 3
3) I'm not sure, but I know that the point that doesn't agree is (0,1).
Now, the answer to this is in the back of the book and they said . . .
"\(\displaystyle \L{}g(x) = \frac{-2x^2 + x}{x}\) and \(\displaystyle \L{}f(x) = -2x +1\)agree except and \(\displaystyle \L{}x=0\)."
My problem is that i don't know how they got that, and it's probably something quite simple, but if someone could explain, I'd be very greatful.
\(\displaystyle \L{}g(x) = \frac{-2x^2 + x}{x}\)
From the graph of this function, I was asked to find three things.
1) \(\displaystyle \L{}\lim_{x \to 0} g(x)\)
2) \(\displaystyle \L{}\lim_{x \to -1} g(x)\)
3) A simpler function that agrees with the given function at all but one point.
So, I looked at the graph and found this:
1) 1
2) 3
3) I'm not sure, but I know that the point that doesn't agree is (0,1).
Now, the answer to this is in the back of the book and they said . . .
"\(\displaystyle \L{}g(x) = \frac{-2x^2 + x}{x}\) and \(\displaystyle \L{}f(x) = -2x +1\)agree except and \(\displaystyle \L{}x=0\)."
My problem is that i don't know how they got that, and it's probably something quite simple, but if someone could explain, I'd be very greatful.
)