Mister Exam
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How to use it?
- Graphing y =:
- x*|x|
-
y=|x-2|-|x+1|+x-2
- x|x|-|x|-6x
- x^2-2x+3
-
Identical expressions
- - zero . five *(abs(\cos(four x+4))+ one / three)/ three arcctg(two x+3)+ one /sqrt(2*x*x- five)
- minus 0.5 multiply by (abs(\ co sinus of e of (4x plus 4)) plus 1 divide by 3) divide by 3arcctg(2x plus 3) plus 1 divide by square root of (2 multiply by x multiply by x minus 5)
- minus zero . five multiply by (abs(\ co sinus of e of (four x plus 4)) plus one divide by three) divide by three arcctg(two x plus 3) plus one divide by square root of (2 multiply by x multiply by x minus five)
- -0.5*(abs(\cos(4x+4))+1/3)/3arcctg(2x+3)+1/√(2*x*x-5)
- -0.5(abs(\cos(4x+4))+1/3)/3arcctg(2x+3)+1/sqrt(2xx-5)
- -0.5abs\cos4x+4+1/3/3arcctg2x+3+1/sqrt2xx-5
- -0.5*(abs(\cos(4x+4))+1 divide by 3) divide by 3arcctg(2x+3)+1 divide by sqrt(2*x*x-5)
-
Similar expressions
- -0.5*(abs(\cos(4x-4))+1/3)/3arcctg(2x+3)+1/sqrt(2*x*x-5)
- -0.5*(abs(\cos(4x+4))+1/3)/3arcctg(2x+3)+1/sqrt(2*x*x+5)
- -0.5*(abs(\cos(4x+4))+1/3)/3arcctg(2x-3)+1/sqrt(2*x*x-5)
- 0.5*(abs(\cos(4x+4))+1/3)/3arcctg(2x+3)+1/sqrt(2*x*x-5)
- -0.5*(abs(\cos(4x+4))+1/3)/3arcctg(2x+3)-1/sqrt(2*x*x-5)
- -0.5*(abs(\cos(4x+4))-1/3)/3arcctg(2x+3)+1/sqrt(2*x*x-5)
Graphing y = -0.5*(abs(\cos(4x+4))+1/3)/3arcctg(2x+3)+1/sqrt(2*x*x-5)
The solution
You have entered
[src]
/-(|cos(4*x + 4)| + 1/3) \
|------------------------|
\ 2 / 1
f(x) = --------------------------*acot(2*x + 3) + -------------
3 ___________
\/ 2*x*x - 5
$$f{\left(x \right)} = \frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}$$
f = ((-(Abs(cos(4*x + 4)) + 1/3)/2)/3)*acot(2*x + 3) + 1/(sqrt(x*(2*x) - 5))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -1.58113883008419$$
$$x_{2} = 1.58113883008419$$
$$x_{1} = -1.58113883008419$$
$$x_{2} = 1.58113883008419$$
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:
$$\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}} = 0$$
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:
$$\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
Vertical asymptotes
Have:
$$x_{1} = -1.58113883008419$$
$$x_{2} = 1.58113883008419$$
$$x_{1} = -1.58113883008419$$
$$x_{2} = 1.58113883008419$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}\right) = \pi \left(- \frac{\left|{\left\langle -1, 1\right\rangle}\right|}{6} - \frac{1}{18}\right)$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \pi \left(- \frac{\left|{\left\langle -1, 1\right\rangle}\right|}{6} - \frac{1}{18}\right)$$
$$\lim_{x \to \infty}\left(\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}\right) = \pi \left(- \frac{\left|{\left\langle -1, 1\right\rangle}\right|}{6} - \frac{1}{18}\right)$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \pi \left(- \frac{\left|{\left\langle -1, 1\right\rangle}\right|}{6} - \frac{1}{18}\right)$$
$$\lim_{x \to \infty}\left(\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}\right) = 0$$
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 ((-(Abs(cos(4*x + 4)) + 1/3)/2)/3)*acot(2*x + 3) + 1/(sqrt((2*x)*x - 5)), divided by x at x->+oo and x ->-oo
Limit on the left could not be calculated
$$\lim_{x \to -\infty}\left(\frac{\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}}{x}\right)$$
Limit on the right could not be calculated
$$\lim_{x \to \infty}\left(\frac{\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}}{x}\right)$$
Limit on the left could not be calculated
$$\lim_{x \to -\infty}\left(\frac{\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}}{x}\right)$$
Limit on the right could not be calculated
$$\lim_{x \to \infty}\left(\frac{\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}}}{x}\right)$$
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:
$$\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}} = - \left(- \frac{\left|{\cos{\left(4 x - 4 \right)}}\right|}{6} - \frac{1}{18}\right) \operatorname{acot}{\left(2 x - 3 \right)} + \frac{1}{\sqrt{2 x^{2} - 5}}$$
- No
$$\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}} = \left(- \frac{\left|{\cos{\left(4 x - 4 \right)}}\right|}{6} - \frac{1}{18}\right) \operatorname{acot}{\left(2 x - 3 \right)} - \frac{1}{\sqrt{2 x^{2} - 5}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}} = - \left(- \frac{\left|{\cos{\left(4 x - 4 \right)}}\right|}{6} - \frac{1}{18}\right) \operatorname{acot}{\left(2 x - 3 \right)} + \frac{1}{\sqrt{2 x^{2} - 5}}$$
- No
$$\frac{\left(-1\right) \frac{1}{2} \left(\left|{\cos{\left(4 x + 4 \right)}}\right| + \frac{1}{3}\right)}{3} \operatorname{acot}{\left(2 x + 3 \right)} + \frac{1}{\sqrt{x 2 x - 5}} = \left(- \frac{\left|{\cos{\left(4 x - 4 \right)}}\right|}{6} - \frac{1}{18}\right) \operatorname{acot}{\left(2 x - 3 \right)} - \frac{1}{\sqrt{2 x^{2} - 5}}$$
- No
so, the function
not is
neither even, nor odd