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Plotting a function graph/ log(0.5,x)

Graphing y = log(0.5,x)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       log(1/2)
f(x) = --------
        log(x)
f(x)=log⁡(12)log⁡(x)f{\left(x \right)} = \frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(x \right)}} 
f = log(1/2)/log(x)
The graph of the function
02468-8-6-4-2-1010-5050
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:
log⁡(12)log⁡(x)=0\frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(x \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(1/2)/log(x).
log⁡(12)log⁡(0)\frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(0 \right)}}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
−log⁡(12)xlog⁡(x)2=0- \frac{\log{\left(\frac{1}{2} \right)}}{x \log{\left(x \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
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
(1+2log⁡(x))log⁡(12)x2log⁡(x)2=0\frac{\left(1 + \frac{2}{\log{\left(x \right)}}\right) \log{\left(\frac{1}{2} \right)}}{x^{2} \log{\left(x \right)}^{2}} = 0
Solve this equation
The roots of this equation
x1=e−2x_{1} = e^{-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:
x1=1x_{1} = 1

lim⁡x→1−((1+2log⁡(x))log⁡(12)x2log⁡(x)2)=∞\lim_{x \to 1^-}\left(\frac{\left(1 + \frac{2}{\log{\left(x \right)}}\right) \log{\left(\frac{1}{2} \right)}}{x^{2} \log{\left(x \right)}^{2}}\right) = \infty
lim⁡x→1+((1+2log⁡(x))log⁡(12)x2log⁡(x)2)=−∞\lim_{x \to 1^+}\left(\frac{\left(1 + \frac{2}{\log{\left(x \right)}}\right) \log{\left(\frac{1}{2} \right)}}{x^{2} \log{\left(x \right)}^{2}}\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−2,∞)\left[e^{-2}, \infty\right)
Convex at the intervals
(−∞,e−2]\left(-\infty, e^{-2}\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→−∞(log⁡(12)log⁡(x))=0\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(x \right)}}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(log⁡(12)log⁡(x))=0\lim_{x \to \infty}\left(\frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(x \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(1/2)/log(x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(log⁡(12)xlog⁡(x))=0\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{1}{2} \right)}}{x \log{\left(x \right)}}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(log⁡(12)xlog⁡(x))=0\lim_{x \to \infty}\left(\frac{\log{\left(\frac{1}{2} \right)}}{x \log{\left(x \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⁡(12)log⁡(x)=log⁡(12)log⁡(−x)\frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(x \right)}} = \frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(- x \right)}}
- No
log⁡(12)log⁡(x)=−log⁡(12)log⁡(−x)\frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(x \right)}} = - \frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(- x \right)}}
- No
so, the function
not is
neither even, nor odd
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