Mister Exam
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-
How to use it?
- Graphing y =:
- 2x^3+3x^2-1
- -2x^2+4x+6
- 2x^3-3x^2
- 2x^2+3x+4
- Derivative of:
-
1/(x+1)^2
- How do you in partial fractions?:
- 1/(x+1)^2
- Integral of d{x}:
-
1/(x+1)^2
-
Identical expressions
- one /(x+ one)^ two
- 1 divide by (x plus 1) squared
- one divide by (x plus one) to the power of two
- 1/(x+1)2
- 1/x+12
- 1/(x+1)²
- 1/(x+1) to the power of 2
- 1/x+1^2
- 1 divide by (x+1)^2
-
Similar expressions
- 1/(x-1)^2
Graphing y = 1/(x+1)^2
The solution
You have entered
[src]
1
f(x) = --------
2
(x + 1)
$$f{\left(x \right)} = \frac{1}{\left(x + 1\right)^{2}}$$
f = 1/((x + 1)^2)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -1$$
$$x_{1} = -1$$
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:
$$\frac{1}{\left(x + 1\right)^{2}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
so we need to solve the equation:
$$\frac{1}{\left(x + 1\right)^{2}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
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
$$\frac{- 2 x - 2}{\left(x + 1\right)^{2} \left(x + 1\right)^{2}} = 0$$
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
$$\frac{- 2 x - 2}{\left(x + 1\right)^{2} \left(x + 1\right)^{2}} = 0$$
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
$$\frac{6}{\left(x + 1\right)^{4}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{6}{\left(x + 1\right)^{4}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = -1$$
$$x_{1} = -1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \frac{1}{\left(x + 1\right)^{2}} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} \frac{1}{\left(x + 1\right)^{2}} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty} \frac{1}{\left(x + 1\right)^{2}} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} \frac{1}{\left(x + 1\right)^{2}} = 0$$
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 + 1)^2), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(x + 1\right)^{2}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{x \left(x + 1\right)^{2}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(x + 1\right)^{2}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{1}{x \left(x + 1\right)^{2}}\right) = 0$$
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:
$$\frac{1}{\left(x + 1\right)^{2}} = \frac{1}{\left(1 - x\right)^{2}}$$
- No
$$\frac{1}{\left(x + 1\right)^{2}} = - \frac{1}{\left(1 - x\right)^{2}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{1}{\left(x + 1\right)^{2}} = \frac{1}{\left(1 - x\right)^{2}}$$
- No
$$\frac{1}{\left(x + 1\right)^{2}} = - \frac{1}{\left(1 - x\right)^{2}}$$
- No
so, the function
not is
neither even, nor odd