Mister Exam
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-
How to use it?
- Graphing y =:
- x^3+3x^2-9x+15
- -x^2+4x
- x^2+4x+2
- x^2-2x-8
- Integral of d{x}:
-
(x+1)/(x^2-4)
-
Identical expressions
- (x+ one)/(x^ two - four)
- (x plus 1) divide by (x squared minus 4)
- (x plus one) divide by (x to the power of two minus four)
- (x+1)/(x2-4)
- x+1/x2-4
- (x+1)/(x²-4)
- (x+1)/(x to the power of 2-4)
- x+1/x^2-4
- (x+1) divide by (x^2-4)
-
Similar expressions
- (x+1)/(x^2+4)
- (x-1)/(x^2-4)
Graphing y = (x+1)/(x^2-4)
The solution
You have entered
[src]
x + 1
f(x) = ------
2
x - 4
$$f{\left(x \right)} = \frac{x + 1}{x^{2} - 4}$$
f = (x + 1)/(x^2 - 4)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{1} = -2$$
$$x_{2} = 2$$
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:
$$\frac{x + 1}{x^{2} - 4} = 0$$
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:
$$\frac{x + 1}{x^{2} - 4} = 0$$
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
$$- \frac{2 x \left(x + 1\right)}{\left(x^{2} - 4\right)^{2}} + \frac{1}{x^{2} - 4} = 0$$
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
$$- \frac{2 x \left(x + 1\right)}{\left(x^{2} - 4\right)^{2}} + \frac{1}{x^{2} - 4} = 0$$
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
$$\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \sqrt[3]{3} - 1 + 3^{\frac{2}{3}}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$\lim_{x \to -2^-}\left(\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}}\right) = -\infty$$
$$\lim_{x \to -2^+}\left(\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = -2$$
- is an inflection point
$$\lim_{x \to 2^-}\left(\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}}\right) = -\infty$$
$$\lim_{x \to 2^+}\left(\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{2} = 2$$
- 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
$$\left(-\infty, - \sqrt[3]{3} - 1 + 3^{\frac{2}{3}}\right]$$
Convex at the intervals
$$\left[- \sqrt[3]{3} - 1 + 3^{\frac{2}{3}}, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \sqrt[3]{3} - 1 + 3^{\frac{2}{3}}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$\lim_{x \to -2^-}\left(\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}}\right) = -\infty$$
$$\lim_{x \to -2^+}\left(\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = -2$$
- is an inflection point
$$\lim_{x \to 2^-}\left(\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}}\right) = -\infty$$
$$\lim_{x \to 2^+}\left(\frac{2 \left(- 2 x + \left(x + 1\right) \left(\frac{4 x^{2}}{x^{2} - 4} - 1\right)\right)}{\left(x^{2} - 4\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{2} = 2$$
- 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
$$\left(-\infty, - \sqrt[3]{3} - 1 + 3^{\frac{2}{3}}\right]$$
Convex at the intervals
$$\left[- \sqrt[3]{3} - 1 + 3^{\frac{2}{3}}, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{1} = -2$$
$$x_{2} = 2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x + 1}{x^{2} - 4}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2} - 4}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty}\left(\frac{x + 1}{x^{2} - 4}\right) = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2} - 4}\right) = 0$$
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 (x + 1)/(x^2 - 4), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{x + 1}{x \left(x^{2} - 4\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{x + 1}{x \left(x^{2} - 4\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{x + 1}{x \left(x^{2} - 4\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{x + 1}{x \left(x^{2} - 4\right)}\right) = 0$$
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:
$$\frac{x + 1}{x^{2} - 4} = \frac{1 - x}{x^{2} - 4}$$
- No
$$\frac{x + 1}{x^{2} - 4} = - \frac{1 - x}{x^{2} - 4}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{x + 1}{x^{2} - 4} = \frac{1 - x}{x^{2} - 4}$$
- No
$$\frac{x + 1}{x^{2} - 4} = - \frac{1 - x}{x^{2} - 4}$$
- No
so, the function
not is
neither even, nor odd