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