Mister Exam
Graphing y = (x+exp(x))/(1+x)
The solution
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x
x + e
f(x) = ------
1 + x
$$f{\left(x \right)} = \frac{x + e^{x}}{x + 1}$$
f = (x + exp(x))/(x + 1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -1$$
$$x_{1} = -1$$
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:
$$\frac{x + e^{x}}{x + 1} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - W\left(1\right)$$
Numerical solution
$$x_{1} = -0.567143290409784$$
so we need to solve the equation:
$$\frac{x + e^{x}}{x + 1} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - W\left(1\right)$$
Numerical solution
$$x_{1} = -0.567143290409784$$
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
$$\frac{e^{x} + 1}{x + 1} - \frac{x + e^{x}}{\left(x + 1\right)^{2}} = 0$$
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
$$\frac{e^{x} + 1}{x + 1} - \frac{x + e^{x}}{\left(x + 1\right)^{2}} = 0$$
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
$$\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -33415.8037497305$$
$$x_{2} = -28330.187841268$$
$$x_{3} = -38501.422170149$$
$$x_{4} = -23244.5760938158$$
$$x_{5} = -15616.1751697251$$
$$x_{6} = -30872.9954039089$$
$$x_{7} = -24092.177650919$$
$$x_{8} = 0.483259389716171$$
$$x_{9} = -37653.818968837$$
$$x_{10} = -19854.1721532085$$
$$x_{11} = -30025.392786181$$
$$x_{12} = -26634.9833453947$$
$$x_{13} = -19006.5718930801$$
$$x_{14} = -29177.7902624869$$
$$x_{15} = -35958.6127125654$$
$$x_{16} = -18158.9720042001$$
$$x_{17} = -34263.4066755697$$
$$x_{18} = -20701.7727389755$$
$$x_{19} = -22396.9747395139$$
$$x_{20} = -13920.98050271$$
$$x_{21} = -32568.2008920951$$
$$x_{22} = -39349.0254160413$$
$$x_{23} = -35111.0096646728$$
$$x_{24} = -41044.2320305184$$
$$x_{25} = -31720.5981081316$$
$$x_{26} = -25787.3812935264$$
$$x_{27} = -21549.3736119485$$
$$x_{28} = -17311.3725411133$$
$$x_{29} = -40196.6287036936$$
$$x_{30} = -41891.8353941377$$
$$x_{31} = -24939.779390143$$
$$x_{32} = -13073.3845123319$$
$$x_{33} = -36806.2158151855$$
$$x_{34} = -16463.7735695997$$
$$x_{35} = -14768.5774399437$$
$$x_{36} = -27482.5855320072$$
$$x_{37} = -12225.7896664061$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1}\right) = \infty$$
$$\lim_{x \to -1^+}\left(\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = -1$$
- 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
$$\left[0.483259389716171, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0.483259389716171\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = -33415.8037497305$$
$$x_{2} = -28330.187841268$$
$$x_{3} = -38501.422170149$$
$$x_{4} = -23244.5760938158$$
$$x_{5} = -15616.1751697251$$
$$x_{6} = -30872.9954039089$$
$$x_{7} = -24092.177650919$$
$$x_{8} = 0.483259389716171$$
$$x_{9} = -37653.818968837$$
$$x_{10} = -19854.1721532085$$
$$x_{11} = -30025.392786181$$
$$x_{12} = -26634.9833453947$$
$$x_{13} = -19006.5718930801$$
$$x_{14} = -29177.7902624869$$
$$x_{15} = -35958.6127125654$$
$$x_{16} = -18158.9720042001$$
$$x_{17} = -34263.4066755697$$
$$x_{18} = -20701.7727389755$$
$$x_{19} = -22396.9747395139$$
$$x_{20} = -13920.98050271$$
$$x_{21} = -32568.2008920951$$
$$x_{22} = -39349.0254160413$$
$$x_{23} = -35111.0096646728$$
$$x_{24} = -41044.2320305184$$
$$x_{25} = -31720.5981081316$$
$$x_{26} = -25787.3812935264$$
$$x_{27} = -21549.3736119485$$
$$x_{28} = -17311.3725411133$$
$$x_{29} = -40196.6287036936$$
$$x_{30} = -41891.8353941377$$
$$x_{31} = -24939.779390143$$
$$x_{32} = -13073.3845123319$$
$$x_{33} = -36806.2158151855$$
$$x_{34} = -16463.7735695997$$
$$x_{35} = -14768.5774399437$$
$$x_{36} = -27482.5855320072$$
$$x_{37} = -12225.7896664061$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1}\right) = \infty$$
$$\lim_{x \to -1^+}\left(\frac{e^{x} - \frac{2 \left(e^{x} + 1\right)}{x + 1} + \frac{2 \left(x + e^{x}\right)}{\left(x + 1\right)^{2}}}{x + 1}\right) = -\infty$$
- the limits are not equal, so
$$x_{1} = -1$$
- 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
$$\left[0.483259389716171, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0.483259389716171\right]$$
Vertical asymptotes
Have:
$$x_{1} = -1$$
$$x_{1} = -1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x + e^{x}}{x + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{x + e^{x}}{x + 1}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{x + e^{x}}{x + 1}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{x + e^{x}}{x + 1}\right) = \infty$$
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 + exp(x))/(1 + x), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{x + e^{x}}{x \left(x + 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{x + e^{x}}{x \left(x + 1\right)}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{x + e^{x}}{x \left(x + 1\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{x + e^{x}}{x \left(x + 1\right)}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
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:
$$\frac{x + e^{x}}{x + 1} = \frac{- x + e^{- x}}{1 - x}$$
- No
$$\frac{x + e^{x}}{x + 1} = - \frac{- x + e^{- x}}{1 - x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{x + e^{x}}{x + 1} = \frac{- x + e^{- x}}{1 - x}$$
- No
$$\frac{x + e^{x}}{x + 1} = - \frac{- x + e^{- x}}{1 - x}$$
- No
so, the function
not is
neither even, nor odd