Mister Exam
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-
How to use it?
- Graphing y =:
- x^3+3x^2-2
-
x^1/3
- y=tan(x+5)
- y=(sqrtx+4)-4
-
Identical expressions
- eight /(log(x- six)/log(three))
- 8 divide by ( logarithm of (x minus 6) divide by logarithm of (3))
- eight divide by ( logarithm of (x minus six) divide by logarithm of (three))
- 8/logx-6/log3
- 8 divide by (log(x-6) divide by log(3))
-
Similar expressions
- 8/(log(x+6)/log(3))
Graphing y = 8/(log(x-6)/log(3))
The solution
You have entered
[src]
8
f(x) = ------------
/log(x - 6)\
|----------|
\ log(3) /
$$f{\left(x \right)} = \frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}}$$
f = 8/((log(x - 6)/log(3)))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 7$$
$$x_{1} = 7$$
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:
$$\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}} = 0$$
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:
$$\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}} = 0$$
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
$$- \frac{8 \log{\left(3 \right)}}{\left(x - 6\right) \log{\left(x - 6 \right)}^{2}} = 0$$
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
$$- \frac{8 \log{\left(3 \right)}}{\left(x - 6\right) \log{\left(x - 6 \right)}^{2}} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{8 \left(1 + \frac{2}{\log{\left(x - 6 \right)}}\right) \log{\left(3 \right)}}{\left(x - 6\right)^{2} \log{\left(x - 6 \right)}^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = e^{-2} + 6$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 7$$
$$\lim_{x \to 7^-}\left(\frac{8 \left(1 + \frac{2}{\log{\left(x - 6 \right)}}\right) \log{\left(3 \right)}}{\left(x - 6\right)^{2} \log{\left(x - 6 \right)}^{2}}\right) = -\infty$$
$$\lim_{x \to 7^+}\left(\frac{8 \left(1 + \frac{2}{\log{\left(x - 6 \right)}}\right) \log{\left(3 \right)}}{\left(x - 6\right)^{2} \log{\left(x - 6 \right)}^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 7$$
- 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
$$\left(-\infty, e^{-2} + 6\right]$$
Convex at the intervals
$$\left[e^{-2} + 6, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{8 \left(1 + \frac{2}{\log{\left(x - 6 \right)}}\right) \log{\left(3 \right)}}{\left(x - 6\right)^{2} \log{\left(x - 6 \right)}^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = e^{-2} + 6$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 7$$
$$\lim_{x \to 7^-}\left(\frac{8 \left(1 + \frac{2}{\log{\left(x - 6 \right)}}\right) \log{\left(3 \right)}}{\left(x - 6\right)^{2} \log{\left(x - 6 \right)}^{2}}\right) = -\infty$$
$$\lim_{x \to 7^+}\left(\frac{8 \left(1 + \frac{2}{\log{\left(x - 6 \right)}}\right) \log{\left(3 \right)}}{\left(x - 6\right)^{2} \log{\left(x - 6 \right)}^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 7$$
- 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
$$\left(-\infty, e^{-2} + 6\right]$$
Convex at the intervals
$$\left[e^{-2} + 6, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = 7$$
$$x_{1} = 7$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}}\right) = 0$$
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 8/((log(x - 6)/log(3))), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{8 \frac{\log{\left(3 \right)}}{\log{\left(x - 6 \right)}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{8 \frac{\log{\left(3 \right)}}{\log{\left(x - 6 \right)}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{8 \frac{\log{\left(3 \right)}}{\log{\left(x - 6 \right)}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{8 \frac{\log{\left(3 \right)}}{\log{\left(x - 6 \right)}}}{x}\right) = 0$$
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:
$$\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}} = \frac{8 \log{\left(3 \right)}}{\log{\left(- x - 6 \right)}}$$
- No
$$\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}} = - \frac{8 \log{\left(3 \right)}}{\log{\left(- x - 6 \right)}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}} = \frac{8 \log{\left(3 \right)}}{\log{\left(- x - 6 \right)}}$$
- No
$$\frac{8}{\frac{1}{\log{\left(3 \right)}} \log{\left(x - 6 \right)}} = - \frac{8 \log{\left(3 \right)}}{\log{\left(- x - 6 \right)}}$$
- No
so, the function
not is
neither even, nor odd