Mister Exam

Other calculators

Plotting a function graph/ (1-x)^(1/x)

Graphing y = (1-x)^(1/x)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
       x _______
f(x) = \/ 1 - x
f(x)=(1−x)1xf{\left(x \right)} = \left(1 - x\right)^{\frac{1}{x}} 
f = (1 - x)^(1/x)
The graph of the function
02468-8-6-4-2-10100.01.0
The domain of the function
The points at which the function is not precisely defined:
x1=0x_{1} = 0
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:
(1−x)1x=0\left(1 - x\right)^{\frac{1}{x}} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=1x_{1} = 1
Numerical solution
x1=1x_{1} = 1
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (1 - x)^(1/x).
(1−0)10\left(1 - 0\right)^{\frac{1}{0}}
The result:
f(0)=NaNf{\left(0 \right)} = \text{NaN}
- the solutions of the equation d'not exist
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\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:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
(1−x)1x(−1x(1−x)−log⁡(1−x)x2)=0\left(1 - x\right)^{\frac{1}{x}} \left(- \frac{1}{x \left(1 - x\right)} - \frac{\log{\left(1 - x \right)}}{x^{2}}\right) = 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
d2dx2f(x)=0\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:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
(1−x)1x(−1(x−1)2+(1x−1−log⁡(1−x)x)2x−2x(x−1)+2log⁡(1−x)x2)x=0\frac{\left(1 - x\right)^{\frac{1}{x}} \left(- \frac{1}{\left(x - 1\right)^{2}} + \frac{\left(\frac{1}{x - 1} - \frac{\log{\left(1 - x \right)}}{x}\right)^{2}}{x} - \frac{2}{x \left(x - 1\right)} + \frac{2 \log{\left(1 - x \right)}}{x^{2}}\right)}{x} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=0x_{1} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(1−x)1x=1\lim_{x \to -\infty} \left(1 - x\right)^{\frac{1}{x}} = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞(1−x)1x=1\lim_{x \to \infty} \left(1 - x\right)^{\frac{1}{x}} = 1
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=1y = 1
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (1 - x)^(1/x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((1−x)1xx)=0\lim_{x \to -\infty}\left(\frac{\left(1 - x\right)^{\frac{1}{x}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((1−x)1xx)=0\lim_{x \to \infty}\left(\frac{\left(1 - x\right)^{\frac{1}{x}}}{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:
(1−x)1x=(x+1)−1x\left(1 - x\right)^{\frac{1}{x}} = \left(x + 1\right)^{- \frac{1}{x}}
- No
(1−x)1x=−(x+1)−1x\left(1 - x\right)^{\frac{1}{x}} = - \left(x + 1\right)^{- \frac{1}{x}}
- No
so, the function
not is
neither even, nor odd
    © Mister Exam – Calculator