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loge(1-(1/x^2))

  • How to use it?

  • Graphing y =:
  • x^3-3*x-2
  • x(2-x)^2
  • -x^2+5x+4
  • 3x^2-4x

  • 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))
Plotting a function graph/ loge(1-(1/x^2))

Graphing y = loge(1-(1/x^2))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          /      1 \
       log|1 - 1*--|
          |       2|
          \      x /
f(x) = -------------
             / 1\   
          log\e /
f(x)=log⁡(1−1⋅1x2)log⁡(e1)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
02468-8-6-4-2-10102.5-2.5
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{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:
log⁡(1−1⋅1x2)log⁡(e1)=0\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
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log(1 - 1/(x^2))/log(exp(1)).
log⁡(1−1⋅102)log⁡(e1)\frac{\log{\left(1 - 1 \cdot \frac{1}{0^{2}} \right)}}{\log{\left(e^{1} \right)}}
The result:
f(0)=∞~f{\left(0 \right)} = \tilde{\infty}
sof doesn't intersect Y
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\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:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
2x3⋅(1−1⋅1x2)log⁡(e1)=0\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
d2dx2f(x)=0\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:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
−2⋅(3+2x2⋅(1−1x2))x4⋅(1−1x2)log⁡(e1)=0- \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
x1=−33x_{1} = - \frac{\sqrt{3}}{3}
x2=33x_{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:
x1=0x_{1} = 0

lim⁡x→0−(−2⋅(3+2x2⋅(1−1x2))x4⋅(1−1x2)log⁡(e1))=∞\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→0+(−2⋅(3+2x2⋅(1−1x2))x4⋅(1−1x2)log⁡(e1))=∞\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
[−33,33]\left[- \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right]
Convex at the intervals
(−∞,−33]∪[33,∞)\left(-\infty, - \frac{\sqrt{3}}{3}\right] \cup \left[\frac{\sqrt{3}}{3}, \infty\right)
Vertical asymptotes
Have:
x1=0x_{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→−∞(log⁡(1−1⋅1x2)log⁡(e1))=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=0y = 0
lim⁡x→∞(log⁡(1−1⋅1x2)log⁡(e1))=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=0y = 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→−∞(log⁡(1−1⋅1x2)xlog⁡(e1))=0\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→∞(log⁡(1−1⋅1x2)xlog⁡(e1))=0\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:
log⁡(1−1⋅1x2)log⁡(e1)=log⁡(1−1⋅1x2)log⁡(e1)\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
log⁡(1−1⋅1x2)log⁡(e1)=−log⁡(1−1⋅1x2)log⁡(e1)\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
Graphing y = loge(1-(1/x^2))
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