Mister Exam
Graphing y = (x^3-6x^2+11x-6)/(x^2-3x+2)
The solution
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3 2
x - 6*x + 11*x - 6
f(x) = --------------------
2
x - 3*x + 2
$$f{\left(x \right)} = \frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2}$$
f = (11*x + x^3 - 6*x^2 - 6)/(x^2 - 3*x + 2)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 1$$
$$x_{2} = 2$$
$$x_{1} = 1$$
$$x_{2} = 2$$
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:
$$\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 3$$
Numerical solution
$$x_{1} = 3$$
so we need to solve the equation:
$$\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 3$$
Numerical solution
$$x_{1} = 3$$
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
$$\frac{\left(3 - 2 x\right) \left(\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6\right)}{\left(\left(x^{2} - 3 x\right) + 2\right)^{2}} + \frac{3 x^{2} - 12 x + 11}{\left(x^{2} - 3 x\right) + 2} = 0$$
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
$$\frac{\left(3 - 2 x\right) \left(\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6\right)}{\left(\left(x^{2} - 3 x\right) + 2\right)^{2}} + \frac{3 x^{2} - 12 x + 11}{\left(x^{2} - 3 x\right) + 2} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Vertical asymptotes
Have:
$$x_{1} = 1$$
$$x_{2} = 2$$
$$x_{1} = 1$$
$$x_{2} = 2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (x^3 - 6*x^2 + 11*x - 6)/(x^2 - 3*x + 2), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{x \left(\left(x^{2} - 3 x\right) + 2\right)}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{x \left(\left(x^{2} - 3 x\right) + 2\right)}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
$$\lim_{x \to -\infty}\left(\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{x \left(\left(x^{2} - 3 x\right) + 2\right)}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{x \left(\left(x^{2} - 3 x\right) + 2\right)}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
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:
$$\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2} = \frac{- x^{3} - 6 x^{2} - 11 x - 6}{x^{2} + 3 x + 2}$$
- No
$$\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2} = - \frac{- x^{3} - 6 x^{2} - 11 x - 6}{x^{2} + 3 x + 2}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2} = \frac{- x^{3} - 6 x^{2} - 11 x - 6}{x^{2} + 3 x + 2}$$
- No
$$\frac{\left(11 x + \left(x^{3} - 6 x^{2}\right)\right) - 6}{\left(x^{2} - 3 x\right) + 2} = - \frac{- x^{3} - 6 x^{2} - 11 x - 6}{x^{2} + 3 x + 2}$$
- No
so, the function
not is
neither even, nor odd