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  • Graphing y =:
  • y=2x^3+9x^2+12x
  • x^3+x^2-x-2
  • (x-3)2
  • x/(2x-3)
  • Identical expressions

  • (exp^x- one)/x
  • ( exponent of to the power of x minus 1) divide by x
  • ( exponent of to the power of x minus one) divide by x
  • (expx-1)/x
  • expx-1/x
  • exp^x-1/x
  • (exp^x-1) divide by x
  • Similar expressions

  • (exp^x+1)/x

Graphing y = (exp^x-1)/x

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
        x    
       E  - 1
f(x) = ------
         x   
$$f{\left(x \right)} = \frac{e^{x} - 1}{x}$$
f = (E^x - 1)/x
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
$$\frac{e^{x} - 1}{x} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (E^x - 1)/x.
$$\frac{-1 + e^{0}}{0}$$
The result:
$$f{\left(0 \right)} = \text{NaN}$$
- the solutions of the equation d'not exist
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$\frac{e^{x}}{x} - \frac{e^{x} - 1}{x^{2}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$\frac{e^{x} - \frac{2 e^{x}}{x} + \frac{2 \left(e^{x} - 1\right)}{x^{2}}}{x} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -17734.1722156161$$
$$x_{2} = -26210.1823018004$$
$$x_{3} = -16038.9742925428$$
$$x_{4} = -41467.0337078697$$
$$x_{5} = -30448.1940837816$$
$$x_{6} = -24514.9784989468$$
$$x_{7} = -31295.7967456477$$
$$x_{8} = -25362.5803223436$$
$$x_{9} = -35533.8111815333$$
$$x_{10} = -19429.3719797754$$
$$x_{11} = -34686.2081625031$$
$$x_{12} = -27057.7844226494$$
$$x_{13} = -33838.605204445$$
$$x_{14} = -15191.3762140647$$
$$x_{15} = -14343.7788634966$$
$$x_{16} = -42314.6370890542$$
$$x_{17} = -32991.0023120587$$
$$x_{18} = -12648.5869326509$$
$$x_{19} = -22819.7753899021$$
$$x_{20} = -38076.6205637343$$
$$x_{21} = -32143.3994905401$$
$$x_{22} = -20276.9724079398$$
$$x_{23} = -21124.5731417223$$
$$x_{24} = -18581.7718990583$$
$$x_{25} = -36381.4142572736$$
$$x_{26} = -23667.3768483792$$
$$x_{27} = -39771.8270548144$$
$$x_{28} = -28752.9890385016$$
$$x_{29} = -16886.5729892827$$
$$x_{30} = -27905.3866720051$$
$$x_{31} = -37229.0173858502$$
$$x_{32} = -11800.9927549711$$
$$x_{33} = -29600.5915120748$$
$$x_{34} = -21972.1741457489$$
$$x_{35} = -38924.2237877047$$
$$x_{36} = -13496.1823781368$$
$$x_{37} = -40619.4303623627$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$

$$\lim_{x \to 0^-}\left(\frac{e^{x} - \frac{2 e^{x}}{x} + \frac{2 \left(e^{x} - 1\right)}{x^{2}}}{x}\right) = \frac{1}{3}$$
$$\lim_{x \to 0^+}\left(\frac{e^{x} - \frac{2 e^{x}}{x} + \frac{2 \left(e^{x} - 1\right)}{x^{2}}}{x}\right) = \frac{1}{3}$$
- limits are equal, then skip the corresponding point

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Have no bends at the whole real axis
Vertical asymptotes
Have:
$$x_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{x} - 1}{x}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{e^{x} - 1}{x}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (E^x - 1)/x, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{x} - 1}{x^{2}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{x} - 1}{x^{2}}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
$$\frac{e^{x} - 1}{x} = - \frac{-1 + e^{- x}}{x}$$
- No
$$\frac{e^{x} - 1}{x} = \frac{-1 + e^{- x}}{x}$$
- No
so, the function
not is
neither even, nor odd