Mister Exam

Other calculators

Plotting a function graph/ 2^(1/x)

Graphing y = 2^(1/x)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
       x ___
f(x) = \/ 2
f(x)=21xf{\left(x \right)} = 2^{\frac{1}{x}} 
f = 2^(1/x)
The graph of the function
0-80-60-40-2020406080-10010001e33
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:
21x=02^{\frac{1}{x}} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2^(1/x).
2102^{\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
−21xlog⁡(2)x2=0- \frac{2^{\frac{1}{x}} \log{\left(2 \right)}}{x^{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
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
21x(2+log⁡(2)x)log⁡(2)x3=0\frac{2^{\frac{1}{x}} \left(2 + \frac{\log{\left(2 \right)}}{x}\right) \log{\left(2 \right)}}{x^{3}} = 0
Solve this equation
The roots of this equation
x1=−log⁡(2)2x_{1} = - \frac{\log{\left(2 \right)}}{2}
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
x1=0x_{1} = 0

lim⁡x→0−(21x(2+log⁡(2)x)log⁡(2)x3)=0\lim_{x \to 0^-}\left(\frac{2^{\frac{1}{x}} \left(2 + \frac{\log{\left(2 \right)}}{x}\right) \log{\left(2 \right)}}{x^{3}}\right) = 0
lim⁡x→0+(21x(2+log⁡(2)x)log⁡(2)x3)=∞\lim_{x \to 0^+}\left(\frac{2^{\frac{1}{x}} \left(2 + \frac{\log{\left(2 \right)}}{x}\right) \log{\left(2 \right)}}{x^{3}}\right) = \infty
- the limits are not equal, so
x1=0x_{1} = 0
- 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
[−log⁡(2)2,∞)\left[- \frac{\log{\left(2 \right)}}{2}, \infty\right)
Convex at the intervals
(−∞,−log⁡(2)2]\left(-\infty, - \frac{\log{\left(2 \right)}}{2}\right]
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→−∞21x=1\lim_{x \to -\infty} 2^{\frac{1}{x}} = 1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=1y = 1
lim⁡x→∞21x=1\lim_{x \to \infty} 2^{\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 2^(1/x), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(21xx)=0\lim_{x \to -\infty}\left(\frac{2^{\frac{1}{x}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(21xx)=0\lim_{x \to \infty}\left(\frac{2^{\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:
21x=2−1x2^{\frac{1}{x}} = 2^{- \frac{1}{x}}
- No
21x=−2−1x2^{\frac{1}{x}} = - 2^{- \frac{1}{x}}
- No
so, the function
not is
neither even, nor odd
    © Mister Exam – Calculator