Mister Exam
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-
How to use it?
- Graphing y =:
-
-x^2+4x-3
- (x^2+5)/(x-3)
- x^3+6x^2+9x
- -6x^2+x+1
-
Identical expressions
- arccos((one -2x)/ three)
- arc co sinus of e of ((1 minus 2x) divide by 3)
- arc co sinus of e of ((one minus 2x) divide by three)
- arccos1-2x/3
- arccos((1-2x) divide by 3)
-
Similar expressions
- arccos((1+2x)/3)
- arccos(((1-2x)/3))
Graphing y = arccos((1-2x)/3)
The solution
You have entered
[src]
/1 - 2*x\
f(x) = acos|-------|
\ 3 /
$$f{\left(x \right)} = \operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)}$$
f = acos(1 - 2*x/3)
The graph of the function
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:
$$\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -1$$
Numerical solution
$$x_{1} = -1$$
so we need to solve the equation:
$$\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -1$$
Numerical solution
$$x_{1} = -1$$
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
$$\frac{2}{3 \sqrt{1 - \frac{\left(1 - 2 x\right)^{2}}{9}}} = 0$$
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
$$\frac{2}{3 \sqrt{1 - \frac{\left(1 - 2 x\right)^{2}}{9}}} = 0$$
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
$$\frac{4 \cdot \left(2 x - 1\right)}{27 \left(1 - \frac{\left(1 - 2 x\right)^{2}}{9}\right)^{\frac{3}{2}}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{1}{2}$$
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
$$\left[\frac{1}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{1}{2}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{4 \cdot \left(2 x - 1\right)}{27 \left(1 - \frac{\left(1 - 2 x\right)^{2}}{9}\right)^{\frac{3}{2}}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{1}{2}$$
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
$$\left[\frac{1}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{1}{2}\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = \infty i$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} \operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = - \infty i$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty} \operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = \infty i$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} \operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = - \infty i$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of acos(1 - 2*x/3), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)}}{x}\right) = 0$$
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:
$$\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = \operatorname{acos}{\left(\frac{2 x}{3} + \frac{1}{3} \right)}$$
- No
$$\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = - \operatorname{acos}{\left(\frac{2 x}{3} + \frac{1}{3} \right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = \operatorname{acos}{\left(\frac{2 x}{3} + \frac{1}{3} \right)}$$
- No
$$\operatorname{acos}{\left(\frac{1 - 2 x}{3} \right)} = - \operatorname{acos}{\left(\frac{2 x}{3} + \frac{1}{3} \right)}$$
- No
so, the function
not is
neither even, nor odd
The graph