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________ f(x) = \/ tan(x)

$$f{\left(x \right)} = \sqrt{\tan{\left(x \right)}}$$

f = sqrt(tan(x))

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:

$$\sqrt{\tan{\left(x \right)}} = 0$$

Solve this equation

The points of intersection with the axis X:

**Analytical solution**

$$x_{1} = 0$$

**Numerical solution**

$$x_{1} = 0$$

so we need to solve the equation:

$$\sqrt{\tan{\left(x \right)}} = 0$$

Solve this equation

The points of intersection with the axis X:

$$x_{1} = 0$$

$$x_{1} = 0$$

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

$$\frac{\frac{\tan^{2}{\left(x \right)}}{2} + \frac{1}{2}}{\sqrt{\tan{\left(x \right)}}} = 0$$

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

$$\frac{\frac{\tan^{2}{\left(x \right)}}{2} + \frac{1}{2}}{\sqrt{\tan{\left(x \right)}}} = 0$$

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

$$\left(4 \sqrt{\tan{\left(x \right)}} - \frac{\tan^{2}{\left(x \right)} + 1}{\tan^{\frac{3}{2}}{\left(x \right)}}\right) \left(\frac{\tan^{2}{\left(x \right)}}{4} + \frac{1}{4}\right) = 0$$

Solve this equation

The roots of this equation

$$x_{1} = - \frac{\pi}{6}$$

$$x_{2} = \frac{\pi}{6}$$

**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

$$\left[\frac{\pi}{6}, \infty\right)$$

Convex at the intervals

$$\left(-\infty, \frac{\pi}{6}\right]$$

$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$

(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

$$\left(4 \sqrt{\tan{\left(x \right)}} - \frac{\tan^{2}{\left(x \right)} + 1}{\tan^{\frac{3}{2}}{\left(x \right)}}\right) \left(\frac{\tan^{2}{\left(x \right)}}{4} + \frac{1}{4}\right) = 0$$

Solve this equation

The roots of this equation

$$x_{1} = - \frac{\pi}{6}$$

$$x_{2} = \frac{\pi}{6}$$

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

$$\left[\frac{\pi}{6}, \infty\right)$$

Convex at the intervals

$$\left(-\infty, \frac{\pi}{6}\right]$$

Horizontal asymptotes

Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo

$$\lim_{x \to -\infty} \sqrt{\tan{\left(x \right)}} = \lim_{x \to -\infty} \sqrt{\tan{\left(x \right)}}$$

Let's take the limit

so,

equation of the horizontal asymptote on the left:

$$y = \lim_{x \to -\infty} \sqrt{\tan{\left(x \right)}}$$

$$\lim_{x \to \infty} \sqrt{\tan{\left(x \right)}} = \lim_{x \to \infty} \sqrt{\tan{\left(x \right)}}$$

Let's take the limit

so,

equation of the horizontal asymptote on the right:

$$y = \lim_{x \to \infty} \sqrt{\tan{\left(x \right)}}$$

$$\lim_{x \to -\infty} \sqrt{\tan{\left(x \right)}} = \lim_{x \to -\infty} \sqrt{\tan{\left(x \right)}}$$

Let's take the limit

so,

equation of the horizontal asymptote on the left:

$$y = \lim_{x \to -\infty} \sqrt{\tan{\left(x \right)}}$$

$$\lim_{x \to \infty} \sqrt{\tan{\left(x \right)}} = \lim_{x \to \infty} \sqrt{\tan{\left(x \right)}}$$

Let's take the limit

so,

equation of the horizontal asymptote on the right:

$$y = \lim_{x \to \infty} \sqrt{\tan{\left(x \right)}}$$

Inclined asymptotes

Inclined asymptote can be found by calculating the limit of sqrt(tan(x)), divided by x at x->+oo and x ->-oo

$$\lim_{x \to -\infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right) = \lim_{x \to -\infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$

Let's take the limit

so,

inclined asymptote equation on the left:

$$y = x \lim_{x \to -\infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$

$$\lim_{x \to \infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right) = \lim_{x \to \infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$

Let's take the limit

so,

inclined asymptote equation on the right:

$$y = x \lim_{x \to \infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$

$$\lim_{x \to -\infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right) = \lim_{x \to -\infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$

Let's take the limit

so,

inclined asymptote equation on the left:

$$y = x \lim_{x \to -\infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$

$$\lim_{x \to \infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right) = \lim_{x \to \infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$

Let's take the limit

so,

inclined asymptote equation on the right:

$$y = x \lim_{x \to \infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$

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:

$$\sqrt{\tan{\left(x \right)}} = \sqrt{- \tan{\left(x \right)}}$$

- No

$$\sqrt{\tan{\left(x \right)}} = - \sqrt{- \tan{\left(x \right)}}$$

- No

so, the function

not is

neither even, nor odd

So, check:

$$\sqrt{\tan{\left(x \right)}} = \sqrt{- \tan{\left(x \right)}}$$

- No

$$\sqrt{\tan{\left(x \right)}} = - \sqrt{- \tan{\left(x \right)}}$$

- No

so, the function

not is

neither even, nor odd

The graph