Mister Exam
Graphing y = (1+(2/(x-5)))^(x+1)
The solution
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x + 1
/ 2 \
f(x) = |1 + -----|
\ x - 5/
$$f{\left(x \right)} = \left(1 + \frac{2}{x - 5}\right)^{x + 1}$$
f = (1 + 2/(x - 5))^(x + 1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 5$$
$$x_{1} = 5$$
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:
$$\left(1 + \frac{2}{x - 5}\right)^{x + 1} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 3$$
Numerical solution
$$x_{1} = 2.99832478747635$$
$$x_{2} = 2.99791643139969$$
so we need to solve the equation:
$$\left(1 + \frac{2}{x - 5}\right)^{x + 1} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = 3$$
Numerical solution
$$x_{1} = 2.99832478747635$$
$$x_{2} = 2.99791643139969$$
Vertical asymptotes
Have:
$$x_{1} = 5$$
$$x_{1} = 5$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \left(1 + \frac{2}{x - 5}\right)^{x + 1} = e^{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = e^{2}$$
$$\lim_{x \to \infty} \left(1 + \frac{2}{x - 5}\right)^{x + 1} = e^{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = e^{2}$$
$$\lim_{x \to -\infty} \left(1 + \frac{2}{x - 5}\right)^{x + 1} = e^{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = e^{2}$$
$$\lim_{x \to \infty} \left(1 + \frac{2}{x - 5}\right)^{x + 1} = e^{2}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = e^{2}$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (1 + 2/(x - 5))^(x + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(1 + \frac{2}{x - 5}\right)^{x + 1}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\left(1 + \frac{2}{x - 5}\right)^{x + 1}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\left(1 + \frac{2}{x - 5}\right)^{x + 1}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\left(1 + \frac{2}{x - 5}\right)^{x + 1}}{x}\right) = 0$$
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:
$$\left(1 + \frac{2}{x - 5}\right)^{x + 1} = \left(1 + \frac{2}{- x - 5}\right)^{1 - x}$$
- No
$$\left(1 + \frac{2}{x - 5}\right)^{x + 1} = - \left(1 + \frac{2}{- x - 5}\right)^{1 - x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\left(1 + \frac{2}{x - 5}\right)^{x + 1} = \left(1 + \frac{2}{- x - 5}\right)^{1 - x}$$
- No
$$\left(1 + \frac{2}{x - 5}\right)^{x + 1} = - \left(1 + \frac{2}{- x - 5}\right)^{1 - x}$$
- No
so, the function
not is
neither even, nor odd