Mister Exam
Graphing y = tg(3x-1)+2tg(x+1)
The solution
You have entered
[src]
f(x) = tan(3*x - 1) + 2*tan(x + 1)
$$f{\left(x \right)} = 2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}$$
f = 2*tan(x + 1) + tan(3*x - 1)
The graph of the function
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
$$2 \tan^{2}{\left(x + 1 \right)} + 3 \tan^{2}{\left(3 x - 1 \right)} + 5 = 0$$
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
$$2 \tan^{2}{\left(x + 1 \right)} + 3 \tan^{2}{\left(3 x - 1 \right)} + 5 = 0$$
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
$$2 \left(2 \left(\tan^{2}{\left(x + 1 \right)} + 1\right) \tan{\left(x + 1 \right)} + 9 \left(\tan^{2}{\left(3 x - 1 \right)} + 1\right) \tan{\left(3 x - 1 \right)}\right) = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$2 \left(2 \left(\tan^{2}{\left(x + 1 \right)} + 1\right) \tan{\left(x + 1 \right)} + 9 \left(\tan^{2}{\left(3 x - 1 \right)} + 1\right) \tan{\left(3 x - 1 \right)}\right) = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \lim_{x \to -\infty}\left(2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}\right)$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \lim_{x \to \infty}\left(2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}\right)$$
True
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \lim_{x \to -\infty}\left(2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}\right)$$
True
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \lim_{x \to \infty}\left(2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}\right)$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of tan(3*x - 1) + 2*tan(x + 1), divided by x at x->+oo and x ->-oo
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}}{x}\right)$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \lim_{x \to \infty}\left(\frac{2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}}{x}\right)$$
True
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}}{x}\right)$$
True
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \lim_{x \to \infty}\left(\frac{2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)}}{x}\right)$$
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:
$$2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)} = - 2 \tan{\left(x - 1 \right)} - \tan{\left(3 x + 1 \right)}$$
- No
$$2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)} = 2 \tan{\left(x - 1 \right)} + \tan{\left(3 x + 1 \right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)} = - 2 \tan{\left(x - 1 \right)} - \tan{\left(3 x + 1 \right)}$$
- No
$$2 \tan{\left(x + 1 \right)} + \tan{\left(3 x - 1 \right)} = 2 \tan{\left(x - 1 \right)} + \tan{\left(3 x + 1 \right)}$$
- No
so, the function
not is
neither even, nor odd