Mister Exam
Graphing y = acot(2*x/5+1)
The solution
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/2*x \
f(x) = acot|--- + 1|
\ 5 /
$$f{\left(x \right)} = \operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)}$$
f = acot((2*x)/5 + 1)
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:
$$\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = 0$$
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:
$$\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = 0$$
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
$$- \frac{2}{5 \left(\left(\frac{2 x}{5} + 1\right)^{2} + 1\right)} = 0$$
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
$$- \frac{2}{5 \left(\left(\frac{2 x}{5} + 1\right)^{2} + 1\right)} = 0$$
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
$$\frac{8 \left(2 x + 5\right)}{125 \left(\frac{\left(2 x + 5\right)^{2}}{25} + 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{5}{2}$$
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
$$\left[- \frac{5}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \frac{5}{2}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{8 \left(2 x + 5\right)}{125 \left(\frac{\left(2 x + 5\right)^{2}}{25} + 1\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{5}{2}$$
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
$$\left[- \frac{5}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \frac{5}{2}\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = \pi$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \pi$$
$$\lim_{x \to \infty} \operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty} \operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = \pi$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \pi$$
$$\lim_{x \to \infty} \operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = 0$$
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 acot((2*x)/5 + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)}}{x}\right) = 0$$
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:
$$\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = - \operatorname{acot}{\left(\frac{2 x}{5} - 1 \right)}$$
- No
$$\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = \operatorname{acot}{\left(\frac{2 x}{5} - 1 \right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = - \operatorname{acot}{\left(\frac{2 x}{5} - 1 \right)}$$
- No
$$\operatorname{acot}{\left(\frac{2 x}{5} + 1 \right)} = \operatorname{acot}{\left(\frac{2 x}{5} - 1 \right)}$$
- No
so, the function
not is
neither even, nor odd