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log(x^2+1)^(1/2)

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  • Similar expressions

  • log(x^2-1)^(1/2)
Plotting a function graph/ log(x^2+1)^(1/2)

Graphing y = log(x^2+1)^(1/2)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          _____________
         /    / 2    \ 
f(x) = \/  log\x  + 1/
f(x)=log⁡(x2+1)f{\left(x \right)} = \sqrt{\log{\left(x^{2} + 1 \right)}} 
f = sqrt(log(x^2 + 1))
The graph of the function
02468-8-6-4-2-10100.02.5
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:
log⁡(x2+1)=0\sqrt{\log{\left(x^{2} + 1 \right)}} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=0x_{1} = 0
Numerical solution
x1=0x_{1} = 0
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to sqrt(log(x^2 + 1)).
log⁡(02+1)\sqrt{\log{\left(0^{2} + 1 \right)}}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
x(x2+1)log⁡(x2+1)=0\frac{x}{\left(x^{2} + 1\right) \sqrt{\log{\left(x^{2} + 1 \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
−2x2x2+1−x2(x2+1)log⁡(x2+1)+1(x2+1)log⁡(x2+1)=0\frac{- \frac{2 x^{2}}{x^{2} + 1} - \frac{x^{2}}{\left(x^{2} + 1\right) \log{\left(x^{2} + 1 \right)}} + 1}{\left(x^{2} + 1\right) \sqrt{\log{\left(x^{2} + 1 \right)}}} = 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→−∞log⁡(x2+1)=∞\lim_{x \to -\infty} \sqrt{\log{\left(x^{2} + 1 \right)}} = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞log⁡(x2+1)=∞\lim_{x \to \infty} \sqrt{\log{\left(x^{2} + 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 sqrt(log(x^2 + 1)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(log⁡(x2+1)x)=0\lim_{x \to -\infty}\left(\frac{\sqrt{\log{\left(x^{2} + 1 \right)}}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(log⁡(x2+1)x)=0\lim_{x \to \infty}\left(\frac{\sqrt{\log{\left(x^{2} + 1 \right)}}}{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:
log⁡(x2+1)=log⁡(x2+1)\sqrt{\log{\left(x^{2} + 1 \right)}} = \sqrt{\log{\left(x^{2} + 1 \right)}}
- Yes
log⁡(x2+1)=−log⁡(x2+1)\sqrt{\log{\left(x^{2} + 1 \right)}} = - \sqrt{\log{\left(x^{2} + 1 \right)}}
- No
so, the function
is
even
The graph
Graphing y = log(x^2+1)^(1/2)
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