Mister Exam
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-
How to use it?
- Graphing y =:
- x^4-2*x^3+1
- x^3-x^2-x-1
- x^4-4x^2+3
- x^3-x^4
-
Identical expressions
- loge(one -(one /x^ two))
- logarithm of e(1 minus (1 divide by x squared ))
- logarithm of e(one minus (one divide by x to the power of two))
- loge(1-(1/x2))
- loge1-1/x2
- loge(1-(1/x²))
- loge(1-(1/x to the power of 2))
- loge1-1/x^2
- loge(1-(1 divide by x^2))
-
Similar expressions
- loge(1+(1/x^2))
Graphing y = loge(1-(1/x^2))
The solution
You have entered
[src]
/ 1 \
log|1 - 1*--|
| 2|
\ x /
f(x) = -------------
/ 1\
log\e /
$$f{\left(x \right)} = \frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}$$
f = log(1 - 1/(x^2))/log(exp(1))
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:
$$\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}} = 0$$
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:
$$\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}} = 0$$
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
$$\frac{2}{x^{3} \cdot \left(1 - 1 \cdot \frac{1}{x^{2}}\right) \log{\left(e^{1} \right)}} = 0$$
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
$$\frac{2}{x^{3} \cdot \left(1 - 1 \cdot \frac{1}{x^{2}}\right) \log{\left(e^{1} \right)}} = 0$$
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
$$- \frac{2 \cdot \left(3 + \frac{2}{x^{2} \cdot \left(1 - \frac{1}{x^{2}}\right)}\right)}{x^{4} \cdot \left(1 - \frac{1}{x^{2}}\right) \log{\left(e^{1} \right)}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{\sqrt{3}}{3}$$
$$x_{2} = \frac{\sqrt{3}}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(- \frac{2 \cdot \left(3 + \frac{2}{x^{2} \cdot \left(1 - \frac{1}{x^{2}}\right)}\right)}{x^{4} \cdot \left(1 - \frac{1}{x^{2}}\right) \log{\left(e^{1} \right)}}\right) = \infty$$
Let's take the limit
$$\lim_{x \to 0^+}\left(- \frac{2 \cdot \left(3 + \frac{2}{x^{2} \cdot \left(1 - \frac{1}{x^{2}}\right)}\right)}{x^{4} \cdot \left(1 - \frac{1}{x^{2}}\right) \log{\left(e^{1} \right)}}\right) = \infty$$
Let's take the limit
- 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:
Concave at the intervals
$$\left[- \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right]$$
Convex at the intervals
$$\left(-\infty, - \frac{\sqrt{3}}{3}\right] \cup \left[\frac{\sqrt{3}}{3}, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$- \frac{2 \cdot \left(3 + \frac{2}{x^{2} \cdot \left(1 - \frac{1}{x^{2}}\right)}\right)}{x^{4} \cdot \left(1 - \frac{1}{x^{2}}\right) \log{\left(e^{1} \right)}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{\sqrt{3}}{3}$$
$$x_{2} = \frac{\sqrt{3}}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(- \frac{2 \cdot \left(3 + \frac{2}{x^{2} \cdot \left(1 - \frac{1}{x^{2}}\right)}\right)}{x^{4} \cdot \left(1 - \frac{1}{x^{2}}\right) \log{\left(e^{1} \right)}}\right) = \infty$$
Let's take the limit
$$\lim_{x \to 0^+}\left(- \frac{2 \cdot \left(3 + \frac{2}{x^{2} \cdot \left(1 - \frac{1}{x^{2}}\right)}\right)}{x^{4} \cdot \left(1 - \frac{1}{x^{2}}\right) \log{\left(e^{1} \right)}}\right) = \infty$$
Let's take the limit
- 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:
Concave at the intervals
$$\left[- \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right]$$
Convex at the intervals
$$\left(-\infty, - \frac{\sqrt{3}}{3}\right] \cup \left[\frac{\sqrt{3}}{3}, \infty\right)$$
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
$$\lim_{x \to -\infty}\left(\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}\right) = 0$$
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 log(1 - 1/(x^2))/log(exp(1)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{x \log{\left(e^{1} \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{x \log{\left(e^{1} \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{x \log{\left(e^{1} \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{x \log{\left(e^{1} \right)}}\right) = 0$$
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:
$$\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}} = \frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}$$
- Yes
$$\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}} = - \frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}$$
- No
so, the function
is
even
So, check:
$$\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}} = \frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}$$
- Yes
$$\frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}} = - \frac{\log{\left(1 - 1 \cdot \frac{1}{x^{2}} \right)}}{\log{\left(e^{1} \right)}}$$
- No
so, the function
is
even
The graph