Mister Exam
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-
How to use it?
- Graphing y =:
-
x^3-3x+1
- 1/4x^4-1\2x^2
- 1/log(x)
-
exp^(1/(x-1))-1
-
Identical expressions
- exp^(one /(x- one))- one
- exponent of to the power of (1 divide by (x minus 1)) minus 1
- exponent of to the power of (one divide by (x minus one)) minus one
- exp(1/(x-1))-1
- exp1/x-1-1
- exp^1/x-1-1
- exp^(1 divide by (x-1))-1
-
Similar expressions
- exp^(1/(x-1))+1
- exp^(1/(x+1))-1
Graphing y = exp^(1/(x-1))-1
The solution
You have entered
[src]
1
1*-----
x - 1
f(x) = e - 1
$$f{\left(x \right)} = e^{1 \cdot \frac{1}{x - 1}} - 1$$
f = E^(1/(x - 1*1)) - 1*1
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 1$$
$$x_{1} = 1$$
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:
$$e^{1 \cdot \frac{1}{x - 1}} - 1 = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
so we need to solve the equation:
$$e^{1 \cdot \frac{1}{x - 1}} - 1 = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
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
$$- \frac{e^{\frac{1}{x - 1}}}{\left(x - 1\right)^{2}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(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
$$- \frac{e^{\frac{1}{x - 1}}}{\left(x - 1\right)^{2}} = 0$$
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
$$\frac{\left(2 + \frac{1}{x - 1}\right) e^{\frac{1}{x - 1}}}{\left(x - 1\right)^{3}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{1}{2}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 1$$
$$\lim_{x \to 1^-}\left(\frac{\left(2 + \frac{1}{x - 1}\right) e^{\frac{1}{x - 1}}}{\left(x - 1\right)^{3}}\right) = 0$$
Let's take the limit
$$\lim_{x \to 1^+}\left(\frac{\left(2 + \frac{1}{x - 1}\right) e^{\frac{1}{x - 1}}}{\left(x - 1\right)^{3}}\right) = \infty$$
Let's take the limit
- the limits are not equal, so
$$x_{1} = 1$$
- is an inflection 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:
Concave at the intervals
$$\left[\frac{1}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{1}{2}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{\left(2 + \frac{1}{x - 1}\right) e^{\frac{1}{x - 1}}}{\left(x - 1\right)^{3}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{1}{2}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 1$$
$$\lim_{x \to 1^-}\left(\frac{\left(2 + \frac{1}{x - 1}\right) e^{\frac{1}{x - 1}}}{\left(x - 1\right)^{3}}\right) = 0$$
Let's take the limit
$$\lim_{x \to 1^+}\left(\frac{\left(2 + \frac{1}{x - 1}\right) e^{\frac{1}{x - 1}}}{\left(x - 1\right)^{3}}\right) = \infty$$
Let's take the limit
- the limits are not equal, so
$$x_{1} = 1$$
- is an inflection 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:
Concave at the intervals
$$\left[\frac{1}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{1}{2}\right]$$
Vertical asymptotes
Have:
$$x_{1} = 1$$
$$x_{1} = 1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(e^{1 \cdot \frac{1}{x - 1}} - 1\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(e^{1 \cdot \frac{1}{x - 1}} - 1\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(e^{1 \cdot \frac{1}{x - 1}} - 1\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(e^{1 \cdot \frac{1}{x - 1}} - 1\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of E^(1/(x - 1*1)) - 1*1, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{1 \cdot \frac{1}{x - 1}} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{1 \cdot \frac{1}{x - 1}} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{e^{1 \cdot \frac{1}{x - 1}} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{1 \cdot \frac{1}{x - 1}} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
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:
$$e^{1 \cdot \frac{1}{x - 1}} - 1 = e^{\frac{1}{- x - 1}} - 1$$
- No
$$e^{1 \cdot \frac{1}{x - 1}} - 1 = - e^{\frac{1}{- x - 1}} + 1$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$e^{1 \cdot \frac{1}{x - 1}} - 1 = e^{\frac{1}{- x - 1}} - 1$$
- No
$$e^{1 \cdot \frac{1}{x - 1}} - 1 = - e^{\frac{1}{- x - 1}} + 1$$
- No
so, the function
not is
neither even, nor odd
The graph