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  • Graphing y =:
  • -x^2-2x+1
  • 5x^2-3x-1
  • -3x+5
  • 3x^2+4x-7

  • Identical expressions

  • one /(log(x)^ three)- one
  • 1 divide by ( logarithm of (x) cubed ) minus 1
  • one divide by ( logarithm of (x) to the power of three) minus one
  • 1/(log(x)3)-1
  • 1/logx3-1
  • 1/(log(x)³)-1
  • 1/(log(x) to the power of 3)-1
  • 1/logx^3-1
  • 1 divide by (log(x)^3)-1

  • Similar expressions

  • 1/(log(x)^3)+1
Plotting a function graph/ 1/(log(x)^3)-1

Graphing y = 1/(log(x)^3)-1

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          1       
f(x) = ------- - 1
          3       
       log (x)
f(x)=−1+1log⁡(x)3f{\left(x \right)} = -1 + \frac{1}{\log{\left(x \right)}^{3}} 
f = -1 + 1/(log(x)^3)
The graph of the function
02468-8-6-4-2-1010-2000020000
The domain of the function
The points at which the function is not precisely defined:
x1=1x_{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:
−1+1log⁡(x)3=0-1 + \frac{1}{\log{\left(x \right)}^{3}} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=ex_{1} = e
Numerical solution
x1=2.71828182845905x_{1} = 2.71828182845905
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 1/(log(x)^3) - 1.
−1+1log⁡(0)3-1 + \frac{1}{\log{\left(0 \right)}^{3}}
The result:
f(0)=−1f{\left(0 \right)} = -1
The point:
(0, -1)
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
−3xlog⁡(x)log⁡(x)3=0- \frac{3}{x \log{\left(x \right)} \log{\left(x \right)}^{3}} = 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
3(1+4log⁡(x))x2log⁡(x)4=0\frac{3 \left(1 + \frac{4}{\log{\left(x \right)}}\right)}{x^{2} \log{\left(x \right)}^{4}} = 0
Solve this equation
The roots of this equation
x1=e−4x_{1} = e^{-4}
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=1x_{1} = 1

lim⁡x→1−(3(1+4log⁡(x))x2log⁡(x)4)=−∞\lim_{x \to 1^-}\left(\frac{3 \left(1 + \frac{4}{\log{\left(x \right)}}\right)}{x^{2} \log{\left(x \right)}^{4}}\right) = -\infty
lim⁡x→1+(3(1+4log⁡(x))x2log⁡(x)4)=∞\lim_{x \to 1^+}\left(\frac{3 \left(1 + \frac{4}{\log{\left(x \right)}}\right)}{x^{2} \log{\left(x \right)}^{4}}\right) = \infty
- the limits are not equal, so
x1=1x_{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
(−∞,e−4]\left(-\infty, e^{-4}\right]
Convex at the intervals
[e−4,∞)\left[e^{-4}, \infty\right)
Vertical asymptotes
Have:
x1=1x_{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→−∞(−1+1log⁡(x)3)=−1\lim_{x \to -\infty}\left(-1 + \frac{1}{\log{\left(x \right)}^{3}}\right) = -1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=−1y = -1
lim⁡x→∞(−1+1log⁡(x)3)=−1\lim_{x \to \infty}\left(-1 + \frac{1}{\log{\left(x \right)}^{3}}\right) = -1
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=−1y = -1
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 1/(log(x)^3) - 1, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(−1+1log⁡(x)3x)=0\lim_{x \to -\infty}\left(\frac{-1 + \frac{1}{\log{\left(x \right)}^{3}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(−1+1log⁡(x)3x)=0\lim_{x \to \infty}\left(\frac{-1 + \frac{1}{\log{\left(x \right)}^{3}}}{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:
−1+1log⁡(x)3=−1+1log⁡(−x)3-1 + \frac{1}{\log{\left(x \right)}^{3}} = -1 + \frac{1}{\log{\left(- x \right)}^{3}}
- No
−1+1log⁡(x)3=1−1log⁡(−x)3-1 + \frac{1}{\log{\left(x \right)}^{3}} = 1 - \frac{1}{\log{\left(- x \right)}^{3}}
- No
so, the function
not is
neither even, nor odd
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