Mister Exam
Graphing y = (1-x)/((x+2)*(x-3))
The solution
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1 - x
f(x) = ---------------
(x + 2)*(x - 3)
$$f{\left(x \right)} = \frac{1 - x}{\left(x - 3\right) \left(x + 2\right)}$$
f = (1 - x)/(((x - 3)*(x + 2)))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -2$$
$$x_{2} = 3$$
$$x_{1} = -2$$
$$x_{2} = 3$$
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:
$$\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 1$$
Numerical solution
$$x_{1} = 1$$
so we need to solve the equation:
$$\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 1$$
Numerical solution
$$x_{1} = 1$$
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
$$\frac{\left(1 - 2 x\right) \left(1 - x\right)}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}} - \frac{1}{\left(x - 3\right) \left(x + 2\right)} = 0$$
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
$$\frac{\left(1 - 2 x\right) \left(1 - x\right)}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}} - \frac{1}{\left(x - 3\right) \left(x + 2\right)} = 0$$
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
$$\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \sqrt[3]{18} + 1 + \sqrt[3]{12}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$x_{2} = 3$$
$$\lim_{x \to -2^-}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = \infty$$
$$\lim_{x \to -2^+}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = -2$$
- is an inflection point
$$\lim_{x \to 3^-}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = \infty$$
$$\lim_{x \to 3^+}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = -\infty$$
- the limits are not equal, so
$$x_{2} = 3$$
- is an inflection point
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
$$\left[- \sqrt[3]{18} + 1 + \sqrt[3]{12}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \sqrt[3]{18} + 1 + \sqrt[3]{12}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \sqrt[3]{18} + 1 + \sqrt[3]{12}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$x_{2} = 3$$
$$\lim_{x \to -2^-}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = \infty$$
$$\lim_{x \to -2^+}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = -2$$
- is an inflection point
$$\lim_{x \to 3^-}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = \infty$$
$$\lim_{x \to 3^+}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = -\infty$$
- the limits are not equal, so
$$x_{2} = 3$$
- is an inflection point
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
$$\left[- \sqrt[3]{18} + 1 + \sqrt[3]{12}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \sqrt[3]{18} + 1 + \sqrt[3]{12}\right]$$
Vertical asymptotes
Have:
$$x_{1} = -2$$
$$x_{2} = 3$$
$$x_{1} = -2$$
$$x_{2} = 3$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (1 - x)/(((x + 2)*(x - 3))), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\frac{1}{\left(x - 3\right) \left(x + 2\right)} \left(1 - x\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\frac{1}{\left(x - 3\right) \left(x + 2\right)} \left(1 - x\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\frac{1}{\left(x - 3\right) \left(x + 2\right)} \left(1 - x\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\frac{1}{\left(x - 3\right) \left(x + 2\right)} \left(1 - x\right)}{x}\right) = 0$$
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:
$$\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = \frac{x + 1}{\left(2 - x\right) \left(- x - 3\right)}$$
- No
$$\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = - \frac{x + 1}{\left(2 - x\right) \left(- x - 3\right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = \frac{x + 1}{\left(2 - x\right) \left(- x - 3\right)}$$
- No
$$\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = - \frac{x + 1}{\left(2 - x\right) \left(- x - 3\right)}$$
- No
so, the function
not is
neither even, nor odd