Mister Exam
Graphing y = ((x-3)(x+4))/(x(x-5))
The solution
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(x - 3)*(x + 4)
f(x) = ---------------
x*(x - 5)
$$f{\left(x \right)} = \frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)}$$
f = ((x - 3)*(x + 4))/((x*(x - 5)))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
$$x_{2} = 5$$
$$x_{1} = 0$$
$$x_{2} = 5$$
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:
$$\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -4$$
$$x_{2} = 3$$
Numerical solution
$$x_{1} = 3$$
$$x_{2} = -4$$
so we need to solve the equation:
$$\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = -4$$
$$x_{2} = 3$$
Numerical solution
$$x_{1} = 3$$
$$x_{2} = -4$$
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
$$\frac{1}{x \left(x - 5\right)} \left(2 x + 1\right) + \frac{\left(5 - 2 x\right) \left(x - 3\right) \left(x + 4\right)}{x^{2} \left(x - 5\right)^{2}} = 0$$
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
$$\frac{1}{x \left(x - 5\right)} \left(2 x + 1\right) + \frac{\left(5 - 2 x\right) \left(x - 3\right) \left(x + 4\right)}{x^{2} \left(x - 5\right)^{2}} = 0$$
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
$$\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \sqrt[3]{12} + 2 + \sqrt[3]{18}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$x_{2} = 5$$
$$\lim_{x \to 0^-}\left(\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)}\right) = -\infty$$
$$\lim_{x \to 0^+}\left(\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- is an inflection point
$$\lim_{x \to 5^-}\left(\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)}\right) = -\infty$$
$$\lim_{x \to 5^+}\left(\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)}\right) = \infty$$
- the limits are not equal, so
$$x_{2} = 5$$
- 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
$$\left(-\infty, - \sqrt[3]{12} + 2 + \sqrt[3]{18}\right]$$
Convex at the intervals
$$\left[- \sqrt[3]{12} + 2 + \sqrt[3]{18}, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \sqrt[3]{12} + 2 + \sqrt[3]{18}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$x_{2} = 5$$
$$\lim_{x \to 0^-}\left(\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)}\right) = -\infty$$
$$\lim_{x \to 0^+}\left(\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 0$$
- is an inflection point
$$\lim_{x \to 5^-}\left(\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)}\right) = -\infty$$
$$\lim_{x \to 5^+}\left(\frac{2 + \frac{\left(x - 3\right) \left(x + 4\right) \left(\left(2 x - 5\right) \left(\frac{1}{x - 5} + \frac{1}{x}\right) - 2 + \frac{2 x - 5}{x - 5} + \frac{2 x - 5}{x}\right)}{x \left(x - 5\right)} - \frac{2 \left(2 x - 5\right) \left(2 x + 1\right)}{x \left(x - 5\right)}}{x \left(x - 5\right)}\right) = \infty$$
- the limits are not equal, so
$$x_{2} = 5$$
- 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
$$\left(-\infty, - \sqrt[3]{12} + 2 + \sqrt[3]{18}\right]$$
Convex at the intervals
$$\left[- \sqrt[3]{12} + 2 + \sqrt[3]{18}, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = 0$$
$$x_{2} = 5$$
$$x_{1} = 0$$
$$x_{2} = 5$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
$$\lim_{x \to -\infty}\left(\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of ((x - 3)*(x + 4))/((x*(x - 5))), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\frac{1}{x \left(x - 5\right)} \left(x - 3\right) \left(x + 4\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\frac{1}{x \left(x - 5\right)} \left(x - 3\right) \left(x + 4\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\frac{1}{x \left(x - 5\right)} \left(x - 3\right) \left(x + 4\right)}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\frac{1}{x \left(x - 5\right)} \left(x - 3\right) \left(x + 4\right)}{x}\right) = 0$$
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:
$$\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)} = - \frac{\left(4 - x\right) \left(- x - 3\right)}{x \left(- x - 5\right)}$$
- No
$$\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)} = \frac{\left(4 - x\right) \left(- x - 3\right)}{x \left(- x - 5\right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)} = - \frac{\left(4 - x\right) \left(- x - 3\right)}{x \left(- x - 5\right)}$$
- No
$$\frac{\left(x - 3\right) \left(x + 4\right)}{x \left(x - 5\right)} = \frac{\left(4 - x\right) \left(- x - 3\right)}{x \left(- x - 5\right)}$$
- No
so, the function
not is
neither even, nor odd