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