Mister Exam
Graphing y = abs(x-(1/x^2))-cos(x)/(log(abs(tan(x)-x)))
The solution
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| 1 | cos(x)
f(x) = |x - --| - -----------------
| 2| log(|tan(x) - x|)
| x |
$$f{\left(x \right)} = \left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}$$
f = |x - 1/x^2| - cos(x)/log(Abs(-x + 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:
$$\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}} = 0$$
Solve this equation
The points of intersection with the axis X:
Numerical solution
$$x_{1} = 1.24431151776631$$
so we need to solve the equation:
$$\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}} = 0$$
Solve this equation
The points of intersection with the axis X:
Numerical solution
$$x_{1} = 1.24431151776631$$
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:
$$y = \lim_{x \to -\infty}\left(\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}\right)$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \lim_{x \to \infty}\left(\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}\right)$$
True
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \lim_{x \to -\infty}\left(\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}\right)$$
True
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \lim_{x \to \infty}\left(\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}\right)$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of |x - 1/x^2| - cos(x)/log(Abs(tan(x) - x)), divided by x at x->+oo and x ->-oo
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}}{x}\right)$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \lim_{x \to \infty}\left(\frac{\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}}{x}\right)$$
True
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}}{x}\right)$$
True
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \lim_{x \to \infty}\left(\frac{\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}}}{x}\right)$$
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:
$$\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}} = \left|{x + \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{x - \tan{\left(x \right)}}\right| \right)}}$$
- No
$$\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}} = - \left|{x + \frac{1}{x^{2}}}\right| + \frac{\cos{\left(x \right)}}{\log{\left(\left|{x - \tan{\left(x \right)}}\right| \right)}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}} = \left|{x + \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{x - \tan{\left(x \right)}}\right| \right)}}$$
- No
$$\left|{x - \frac{1}{x^{2}}}\right| - \frac{\cos{\left(x \right)}}{\log{\left(\left|{- x + \tan{\left(x \right)}}\right| \right)}} = - \left|{x + \frac{1}{x^{2}}}\right| + \frac{\cos{\left(x \right)}}{\log{\left(\left|{x - \tan{\left(x \right)}}\right| \right)}}$$
- No
so, the function
not is
neither even, nor odd