Mister Exam

Graphing y = sqrt(tan(x))

Function f()

The graph:

from to

Intersection points:

does show?

Enter:

The solution

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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
$$x_{1} = 0$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to sqrt(tan(x)).
$$\sqrt{\tan{\left(0 \right)}}$$
The result:
$$f{\left(0 \right)} = 0$$
The point:
(0, 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
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
$$\left(- \frac{\tan^{2}{\left(x \right)} + 1}{\tan^{\frac{3}{2}}{\left(x \right)}} + 4 \sqrt{\tan{\left(x \right)}}\right) \left(\frac{\tan^{2}{\left(x \right)}}{4} + \frac{1}{4}\right) = 0$$
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
$$\left(-\infty, \frac{\pi}{6}\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
True

Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \lim_{x \to -\infty} \sqrt{\tan{\left(x \right)}}$$
True

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
True

Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{\sqrt{\tan{\left(x \right)}}}{x}\right)$$
True

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
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