Mister Exam
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|x^2+x-2|
e^(1/x)
e^(1/x)
Graphing y = e^(1/x)
The solution
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
$$x_{1} = 0$$
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:
$$e^{\frac{1}{x}} = 0$$
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:
$$e^{\frac{1}{x}} = 0$$
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
$$- \frac{e^{\frac{1}{x}}}{x^{2}} = 0$$
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
$$- \frac{e^{\frac{1}{x}}}{x^{2}} = 0$$
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
$$\frac{\left(2 + \frac{1}{x}\right) e^{\frac{1}{x}}}{x^{3}} = 0$$
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:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{\left(2 + \frac{1}{x}\right) e^{\frac{1}{x}}}{x^{3}}\right) = 0$$
$$\lim_{x \to 0^+}\left(\frac{\left(2 + \frac{1}{x}\right) e^{\frac{1}{x}}}{x^{3}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- 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
$$\frac{\left(2 + \frac{1}{x}\right) e^{\frac{1}{x}}}{x^{3}} = 0$$
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:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{\left(2 + \frac{1}{x}\right) e^{\frac{1}{x}}}{x^{3}}\right) = 0$$
$$\lim_{x \to 0^+}\left(\frac{\left(2 + \frac{1}{x}\right) e^{\frac{1}{x}}}{x^{3}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- 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:
$$x_{1} = 0$$
$$x_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} e^{\frac{1}{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty} e^{\frac{1}{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty} e^{\frac{1}{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty} e^{\frac{1}{x}} = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of E^(1/x), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{\frac{1}{x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{\frac{1}{x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{e^{\frac{1}{x}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{\frac{1}{x}}}{x}\right) = 0$$
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:
$$e^{\frac{1}{x}} = e^{- \frac{1}{x}}$$
- No
$$e^{\frac{1}{x}} = - e^{- \frac{1}{x}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$e^{\frac{1}{x}} = e^{- \frac{1}{x}}$$
- No
$$e^{\frac{1}{x}} = - e^{- \frac{1}{x}}$$
- No
so, the function
not is
neither even, nor odd