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x/(x^2+2*x-3)
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  • Graphing y =:
  • |x^2-x-2| |x^2-x-2|
  • -x²+6x-5
  • x^2-8x
  • x^2+5x-4
  • How do you in partial fractions?:
  • x/(x^2+2*x-3)
  • Integral of d{x}:
  • x/(x^2+2*x-3) x/(x^2+2*x-3)
  • Identical expressions

  • x/(x^ two + two *x- three)
  • x divide by (x squared plus 2 multiply by x minus 3)
  • x divide by (x to the power of two plus two multiply by x minus three)
  • x/(x2+2*x-3)
  • x/x2+2*x-3
  • x/(x²+2*x-3)
  • x/(x to the power of 2+2*x-3)
  • x/(x^2+2x-3)
  • x/(x2+2x-3)
  • x/x2+2x-3
  • x/x^2+2x-3
  • x divide by (x^2+2*x-3)
  • Similar expressions

  • x/(x^2+2*x+3)
  • x/(x^2-2*x-3)

Graphing y = x/(x^2+2*x-3)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
            x      
f(x) = ------------
        2          
       x  + 2*x - 3
f(x)=xx2+2x3f{\left(x \right)} = \frac{x}{x^{2} + 2 x - 3}
f = x/(x^2 + 2*x - 1*3)
The graph of the function
02468-8-6-4-2-1010-5050
The domain of the function
The points at which the function is not precisely defined:
x1=3x_{1} = -3
x2=1x_{2} = 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:
xx2+2x3=0\frac{x}{x^{2} + 2 x - 3} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=0x_{1} = 0
Numerical solution
x1=0x_{1} = 0
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to x/(x^2 + 2*x - 1*3).
0(1)3+02+20\frac{0}{\left(-1\right) 3 + 0^{2} + 2 \cdot 0}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\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:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
x(2x2)(x2+2x3)2+1x2+2x3=0\frac{x \left(- 2 x - 2\right)}{\left(x^{2} + 2 x - 3\right)^{2}} + \frac{1}{x^{2} + 2 x - 3} = 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
d2dx2f(x)=0\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:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2(x(4(x+1)2x2+2x31)2x2)(x2+2x3)2=0\frac{2 \left(x \left(\frac{4 \left(x + 1\right)^{2}}{x^{2} + 2 x - 3} - 1\right) - 2 x - 2\right)}{\left(x^{2} + 2 x - 3\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=323+33x_{1} = - 3^{\frac{2}{3}} + \sqrt[3]{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:
x1=3x_{1} = -3
x2=1x_{2} = 1

limx3(2(x(4(x+1)2x2+2x31)2x2)(x2+2x3)2)=\lim_{x \to -3^-}\left(\frac{2 \left(x \left(\frac{4 \left(x + 1\right)^{2}}{x^{2} + 2 x - 3} - 1\right) - 2 x - 2\right)}{\left(x^{2} + 2 x - 3\right)^{2}}\right) = -\infty
Let's take the limit
limx3+(2(x(4(x+1)2x2+2x31)2x2)(x2+2x3)2)=\lim_{x \to -3^+}\left(\frac{2 \left(x \left(\frac{4 \left(x + 1\right)^{2}}{x^{2} + 2 x - 3} - 1\right) - 2 x - 2\right)}{\left(x^{2} + 2 x - 3\right)^{2}}\right) = \infty
Let's take the limit
- the limits are not equal, so
x1=3x_{1} = -3
- is an inflection point
limx1(2(x(4(x+1)2x2+2x31)2x2)(x2+2x3)2)=\lim_{x \to 1^-}\left(\frac{2 \left(x \left(\frac{4 \left(x + 1\right)^{2}}{x^{2} + 2 x - 3} - 1\right) - 2 x - 2\right)}{\left(x^{2} + 2 x - 3\right)^{2}}\right) = -\infty
Let's take the limit
limx1+(2(x(4(x+1)2x2+2x31)2x2)(x2+2x3)2)=\lim_{x \to 1^+}\left(\frac{2 \left(x \left(\frac{4 \left(x + 1\right)^{2}}{x^{2} + 2 x - 3} - 1\right) - 2 x - 2\right)}{\left(x^{2} + 2 x - 3\right)^{2}}\right) = \infty
Let's take the limit
- the limits are not equal, so
x2=1x_{2} = 1
- 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
(,323+33]\left(-\infty, - 3^{\frac{2}{3}} + \sqrt[3]{3}\right]
Convex at the intervals
[323+33,)\left[- 3^{\frac{2}{3}} + \sqrt[3]{3}, \infty\right)
Vertical asymptotes
Have:
x1=3x_{1} = -3
x2=1x_{2} = 1
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
limx(xx2+2x3)=0\lim_{x \to -\infty}\left(\frac{x}{x^{2} + 2 x - 3}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
limx(xx2+2x3)=0\lim_{x \to \infty}\left(\frac{x}{x^{2} + 2 x - 3}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=0y = 0
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of x/(x^2 + 2*x - 1*3), divided by x at x->+oo and x ->-oo
limx1x2+2x3=0\lim_{x \to -\infty} \frac{1}{x^{2} + 2 x - 3} = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
limx1x2+2x3=0\lim_{x \to \infty} \frac{1}{x^{2} + 2 x - 3} = 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:
xx2+2x3=xx22x3\frac{x}{x^{2} + 2 x - 3} = - \frac{x}{x^{2} - 2 x - 3}
- No
xx2+2x3=xx22x3\frac{x}{x^{2} + 2 x - 3} = \frac{x}{x^{2} - 2 x - 3}
- No
so, the function
not is
neither even, nor odd
The graph
Graphing y = x/(x^2+2*x-3)