Mister Exam
Graphing y = tan(arcsin(x)+1)
The solution
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f(x) = tan(asin(x) + 1)
$$f{\left(x \right)} = \tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)}$$
f = tan(asin(x) + 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:
$$\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \sin{\left(1 \right)}$$
Numerical solution
$$x_{1} = -0.841470984807897$$
so we need to solve the equation:
$$\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \sin{\left(1 \right)}$$
Numerical solution
$$x_{1} = -0.841470984807897$$
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
$$\frac{\tan^{2}{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} + 1}{\sqrt{1 - x^{2}}} = 0$$
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
$$\frac{\tan^{2}{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} + 1}{\sqrt{1 - x^{2}}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = \tan{\left(1 + \infty i \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \tan{\left(1 + \infty i \right)}$$
$$\lim_{x \to \infty} \tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = \tan{\left(1 - \infty i \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \tan{\left(1 - \infty i \right)}$$
$$\lim_{x \to -\infty} \tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = \tan{\left(1 + \infty i \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \tan{\left(1 + \infty i \right)}$$
$$\lim_{x \to \infty} \tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = \tan{\left(1 - \infty i \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \tan{\left(1 - \infty i \right)}$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of tan(asin(x) + 1), divided by x at x->+oo and x ->-oo
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)}}{x}\right)$$
$$\lim_{x \to \infty}\left(\frac{\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
True
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)}}{x}\right)$$
$$\lim_{x \to \infty}\left(\frac{\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)}}{x}\right) = 0$$
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:
$$\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = - \tan{\left(\operatorname{asin}{\left(x \right)} - 1 \right)}$$
- No
$$\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = \tan{\left(\operatorname{asin}{\left(x \right)} - 1 \right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = - \tan{\left(\operatorname{asin}{\left(x \right)} - 1 \right)}$$
- No
$$\tan{\left(\operatorname{asin}{\left(x \right)} + 1 \right)} = \tan{\left(\operatorname{asin}{\left(x \right)} - 1 \right)}$$
- No
so, the function
not is
neither even, nor odd