Mister Exam
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-
How to use it?
- Graphing y =:
- -x^2+4x
- (x^2-4x+1)/(x-4)
- -x^2+2x+4
- -x^2+2x-3
-
Identical expressions
- two ln(x/(x-2))- one
- 2ln(x divide by (x minus 2)) minus 1
- two ln(x divide by (x minus 2)) minus one
- 2lnx/x-2-1
- 2ln(x divide by (x-2))-1
-
Similar expressions
- 2lnx/(x-2)-1
- 2ln(x/(x-2))+1
- 2ln(x/(x+2))-1
Graphing y = 2ln(x/(x-2))-1
The solution
You have entered
[src]
/ x \
f(x) = 2*log|-----| - 1
\x - 2/
$$f{\left(x \right)} = 2 \log{\left(\frac{x}{x - 2} \right)} - 1$$
f = 2*log(x/(x - 1*2)) - 1*1
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 2$$
$$x_{1} = 2$$
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:
$$2 \log{\left(\frac{x}{x - 2} \right)} - 1 = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \frac{2 e^{\frac{1}{2}}}{1 - e^{\frac{1}{2}}}$$
Numerical solution
$$x_{1} = 5.0829881650736$$
so we need to solve the equation:
$$2 \log{\left(\frac{x}{x - 2} \right)} - 1 = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \frac{2 e^{\frac{1}{2}}}{1 - e^{\frac{1}{2}}}$$
Numerical solution
$$x_{1} = 5.0829881650736$$
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
$$\frac{2 \left(x - 2\right) \left(- \frac{x}{\left(x - 2\right)^{2}} + \frac{1}{x - 2}\right)}{x} = 0$$
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
$$\frac{2 \left(x - 2\right) \left(- \frac{x}{\left(x - 2\right)^{2}} + \frac{1}{x - 2}\right)}{x} = 0$$
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
$$\frac{2 \left(\frac{x}{x - 2} - 1\right) \left(\frac{1}{x - 2} + \frac{1}{x}\right)}{x} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 1$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 2$$
$$\lim_{x \to 2^-}\left(\frac{2 \left(\frac{x}{x - 2} - 1\right) \left(\frac{1}{x - 2} + \frac{1}{x}\right)}{x}\right) = \infty$$
Let's take the limit
$$\lim_{x \to 2^+}\left(\frac{2 \left(\frac{x}{x - 2} - 1\right) \left(\frac{1}{x - 2} + \frac{1}{x}\right)}{x}\right) = \infty$$
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
$$\left[1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 1\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{2 \left(\frac{x}{x - 2} - 1\right) \left(\frac{1}{x - 2} + \frac{1}{x}\right)}{x} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 1$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 2$$
$$\lim_{x \to 2^-}\left(\frac{2 \left(\frac{x}{x - 2} - 1\right) \left(\frac{1}{x - 2} + \frac{1}{x}\right)}{x}\right) = \infty$$
Let's take the limit
$$\lim_{x \to 2^+}\left(\frac{2 \left(\frac{x}{x - 2} - 1\right) \left(\frac{1}{x - 2} + \frac{1}{x}\right)}{x}\right) = \infty$$
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
$$\left[1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 1\right]$$
Vertical asymptotes
Have:
$$x_{1} = 2$$
$$x_{1} = 2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(2 \log{\left(\frac{x}{x - 2} \right)} - 1\right) = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -1$$
$$\lim_{x \to \infty}\left(2 \log{\left(\frac{x}{x - 2} \right)} - 1\right) = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = -1$$
$$\lim_{x \to -\infty}\left(2 \log{\left(\frac{x}{x - 2} \right)} - 1\right) = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -1$$
$$\lim_{x \to \infty}\left(2 \log{\left(\frac{x}{x - 2} \right)} - 1\right) = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = -1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2*log(x/(x - 1*2)) - 1*1, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{2 \log{\left(\frac{x}{x - 2} \right)} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{2 \log{\left(\frac{x}{x - 2} \right)} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{2 \log{\left(\frac{x}{x - 2} \right)} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{2 \log{\left(\frac{x}{x - 2} \right)} - 1}{x}\right) = 0$$
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:
$$2 \log{\left(\frac{x}{x - 2} \right)} - 1 = 2 \log{\left(- \frac{x}{- x - 2} \right)} - 1$$
- No
$$2 \log{\left(\frac{x}{x - 2} \right)} - 1 = 1 - 2 \log{\left(- \frac{x}{- x - 2} \right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$2 \log{\left(\frac{x}{x - 2} \right)} - 1 = 2 \log{\left(- \frac{x}{- x - 2} \right)} - 1$$
- No
$$2 \log{\left(\frac{x}{x - 2} \right)} - 1 = 1 - 2 \log{\left(- \frac{x}{- x - 2} \right)}$$
- No
so, the function
not is
neither even, nor odd
The graph