What are the asymptotes and removable discontinuities, if any, of $f (x) = \frac{e^{x}}{1 - e^{3 x^{2} - x}}$ $f(x)=e^x/(1-e^(3x^2-x))$ ?

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Answer 1 · 2018-03-01T14:23:06

No discontinuities.

Vertical asymptotes at $x = 0$ $x=0$ and $x = \frac{1}{3}$ $x=1/3$

Horizontal asymptote at $y = 0$ $y=0$

To find the vertical asymptotes, we equate the denominator to $0$ $0$ .

Here,

$1 - e^{3 x^{2} - x} = 0$ $1-e^(3x^2-x)=0$

$- e^{3 x^{2} - x} = - 1$ $-e^(3x^2-x)=-1$

$e^{3 x^{2} - x} = 1$ $e^(3x^2-x)=1$

$ln (e^{3 x^{2} - x}) = ln (1)$ $ln(e^(3x^2-x))=ln(1)$

$3 x^{2} - x = 0$ $3x^2-x=0$

$x (3 x - 1) = 0$ $x(3x-1)=0$

$x = 0, 3 x - 1 = 0$ $x=0, 3x-1=0$

$x = 0, x = \frac{1}{3}$ $x=0,x=1/3$

$x = \frac{1}{3}, 0$ $x=1/3,0$

So we find vertical asymptote is at $x = \frac{1}{3}, 0$ $x=1/3,0$

To find the horizontal asymptote, we must know one crucial fact: all exponential functions have horizontal asymptotes at $y = 0$ $y=0$

Obviously, the graphs of $k^{x} + n$ $k^x+n$ and other such graphs do not count.

Graphing:

graph{(e^x)/(1-e^(3x^2-x)) [-18.02, 18.03, -9.01, 9.01]}

What are the asymptotes and removable discontinuities, if any, of f(x)=e^x/(1-e^(3x^2-x))f(x)=ex1−e3x2−xf(x)=e^x/(1-e^(3x^2-x))?