Mister Exam
Graphing y = (x+2)exp^1-x
The solution
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1 f(x) = (x + 2)*E - x
$$f{\left(x \right)} = - x + e^{1} \left(x + 2\right)$$
f = -x + E^1*(x + 2)
The graph of the function
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:
$$- x + e^{1} \left(x + 2\right) = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = \frac{2 e}{1 - e}$$
Numerical solution
$$x_{1} = -3.16395341373865$$
so we need to solve the equation:
$$- x + e^{1} \left(x + 2\right) = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = \frac{2 e}{1 - e}$$
Numerical solution
$$x_{1} = -3.16395341373865$$
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
$$-1 + e^{1} = 0$$
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
$$-1 + e^{1} = 0$$
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
$$0 = 0$$
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
$$0 = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(- x + e^{1} \left(x + 2\right)\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(- x + e^{1} \left(x + 2\right)\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(- x + e^{1} \left(x + 2\right)\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(- x + e^{1} \left(x + 2\right)\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (x + 2)*E^1 - x, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{- x + e^{1} \left(x + 2\right)}{x}\right) = -1 + e$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \left(-1 + e\right)$$
$$\lim_{x \to \infty}\left(\frac{- x + e^{1} \left(x + 2\right)}{x}\right) = -1 + e$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \left(-1 + e\right)$$
$$\lim_{x \to -\infty}\left(\frac{- x + e^{1} \left(x + 2\right)}{x}\right) = -1 + e$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \left(-1 + e\right)$$
$$\lim_{x \to \infty}\left(\frac{- x + e^{1} \left(x + 2\right)}{x}\right) = -1 + e$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \left(-1 + e\right)$$
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:
$$- x + e^{1} \left(x + 2\right) = x + e \left(2 - x\right)$$
- No
$$- x + e^{1} \left(x + 2\right) = - x - e \left(2 - x\right)$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$- x + e^{1} \left(x + 2\right) = x + e \left(2 - x\right)$$
- No
$$- x + e^{1} \left(x + 2\right) = - x - e \left(2 - x\right)$$
- No
so, the function
not is
neither even, nor odd