Mister Exam

Other calculators

Graphing y = sqrt((5-x)/(x-2))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
           _______
          / 5 - x 
f(x) =   /  ----- 
       \/   x - 2 
$$f{\left(x \right)} = \sqrt{\frac{5 - x}{x - 2}}$$
f = sqrt((5 - x)/(x - 2))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 2$$
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:
$$\sqrt{\frac{5 - x}{x - 2}} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 5$$
Numerical solution
$$x_{1} = 5$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to sqrt((5 - x)/(x - 2)).
$$\sqrt{\frac{5 - 0}{-2}}$$
The result:
$$f{\left(0 \right)} = \frac{\sqrt{10} i}{2}$$
The point:
(0, i*sqrt(10)/2)
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
$$\frac{\sqrt{\frac{5 - x}{x - 2}} \left(x - 2\right) \left(- \frac{5 - x}{2 \left(x - 2\right)^{2}} - \frac{1}{2 \left(x - 2\right)}\right)}{5 - x} = 0$$
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
$$\frac{\sqrt{- \frac{x - 5}{x - 2}} \left(\frac{x - 5}{x - 2} - 1\right) \left(\frac{2}{x - 2} + \frac{\frac{x - 5}{x - 2} - 1}{x - 5} + \frac{2}{x - 5}\right)}{4 \left(x - 5\right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{17}{4}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 2$$

$$\lim_{x \to 2^-}\left(\frac{\sqrt{- \frac{x - 5}{x - 2}} \left(\frac{x - 5}{x - 2} - 1\right) \left(\frac{2}{x - 2} + \frac{\frac{x - 5}{x - 2} - 1}{x - 5} + \frac{2}{x - 5}\right)}{4 \left(x - 5\right)}\right) = \infty i$$
$$\lim_{x \to 2^+}\left(\frac{\sqrt{- \frac{x - 5}{x - 2}} \left(\frac{x - 5}{x - 2} - 1\right) \left(\frac{2}{x - 2} + \frac{\frac{x - 5}{x - 2} - 1}{x - 5} + \frac{2}{x - 5}\right)}{4 \left(x - 5\right)}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 2$$
- 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
$$\left(-\infty, \frac{17}{4}\right]$$
Convex at the intervals
$$\left[\frac{17}{4}, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = 2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \sqrt{\frac{5 - x}{x - 2}} = i$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = i$$
$$\lim_{x \to \infty} \sqrt{\frac{5 - x}{x - 2}} = i$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = i$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sqrt((5 - x)/(x - 2)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sqrt{\frac{5 - x}{x - 2}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{\frac{5 - x}{x - 2}}}{x}\right) = 0$$
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:
$$\sqrt{\frac{5 - x}{x - 2}} = \sqrt{\frac{x + 5}{- x - 2}}$$
- No
$$\sqrt{\frac{5 - x}{x - 2}} = - \sqrt{\frac{x + 5}{- x - 2}}$$
- No
so, the function
not is
neither even, nor odd