Mister Exam
Graphing y = acos((2*x)/(1+x))
The solution
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/ 2*x \
f(x) = acos|-----|
\1 + x/
$$f{\left(x \right)} = \operatorname{acos}{\left(\frac{2 x}{x + 1} \right)}$$
f = acos((2*x)/(x + 1))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -1$$
$$x_{1} = -1$$
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:
$$\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = 0$$
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:
$$\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = 0$$
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
$$- \frac{- \frac{2 x}{\left(x + 1\right)^{2}} + \frac{2}{x + 1}}{\sqrt{- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1}} = 0$$
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
$$- \frac{- \frac{2 x}{\left(x + 1\right)^{2}} + \frac{2}{x + 1}}{\sqrt{- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1}} = 0$$
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
$$- \frac{4 \left(\frac{x}{x + 1} - 1\right) \left(\frac{2 x \left(\frac{x}{x + 1} - 1\right)}{\left(x + 1\right) \left(- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{\sqrt{3}}{3}$$
$$x_{2} = \frac{\sqrt{3}}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(- \frac{4 \left(\frac{x}{x + 1} - 1\right) \left(\frac{2 x \left(\frac{x}{x + 1} - 1\right)}{\left(x + 1\right) \left(- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1}}\right) = \infty i$$
$$\lim_{x \to -1^+}\left(- \frac{4 \left(\frac{x}{x + 1} - 1\right) \left(\frac{2 x \left(\frac{x}{x + 1} - 1\right)}{\left(x + 1\right) \left(- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1}}\right) = - \infty i$$
- the limits are not equal, so
$$x_{1} = -1$$
- 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
$$\left(-\infty, \frac{\sqrt{3}}{3}\right]$$
Convex at the intervals
$$\left[\frac{\sqrt{3}}{3}, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$- \frac{4 \left(\frac{x}{x + 1} - 1\right) \left(\frac{2 x \left(\frac{x}{x + 1} - 1\right)}{\left(x + 1\right) \left(- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{\sqrt{3}}{3}$$
$$x_{2} = \frac{\sqrt{3}}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(- \frac{4 \left(\frac{x}{x + 1} - 1\right) \left(\frac{2 x \left(\frac{x}{x + 1} - 1\right)}{\left(x + 1\right) \left(- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1}}\right) = \infty i$$
$$\lim_{x \to -1^+}\left(- \frac{4 \left(\frac{x}{x + 1} - 1\right) \left(\frac{2 x \left(\frac{x}{x + 1} - 1\right)}{\left(x + 1\right) \left(- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1\right)} + 1\right)}{\left(x + 1\right)^{2} \sqrt{- \frac{4 x^{2}}{\left(x + 1\right)^{2}} + 1}}\right) = - \infty i$$
- the limits are not equal, so
$$x_{1} = -1$$
- 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
$$\left(-\infty, \frac{\sqrt{3}}{3}\right]$$
Convex at the intervals
$$\left[\frac{\sqrt{3}}{3}, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = -1$$
$$x_{1} = -1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = i \log{\left(\sqrt{3} + 2 \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = i \log{\left(\sqrt{3} + 2 \right)}$$
$$\lim_{x \to \infty} \operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = i \log{\left(\sqrt{3} + 2 \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = i \log{\left(\sqrt{3} + 2 \right)}$$
$$\lim_{x \to -\infty} \operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = i \log{\left(\sqrt{3} + 2 \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = i \log{\left(\sqrt{3} + 2 \right)}$$
$$\lim_{x \to \infty} \operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = i \log{\left(\sqrt{3} + 2 \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = i \log{\left(\sqrt{3} + 2 \right)}$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of acos((2*x)/(1 + x)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)}}{x}\right) = 0$$
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:
$$\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = \operatorname{acos}{\left(- \frac{2 x}{1 - x} \right)}$$
- No
$$\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = - \operatorname{acos}{\left(- \frac{2 x}{1 - x} \right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = \operatorname{acos}{\left(- \frac{2 x}{1 - x} \right)}$$
- No
$$\operatorname{acos}{\left(\frac{2 x}{x + 1} \right)} = - \operatorname{acos}{\left(- \frac{2 x}{1 - x} \right)}$$
- No
so, the function
not is
neither even, nor odd