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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
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f(x) = log\x + \/  1 + x  /
f(x)=log⁡(x+x2+1)f{\left(x \right)} = \log{\left(x + \sqrt{x^{2} + 1} \right)} 
f = log(x + sqrt(x^2 + 1))
The graph of the function
02468-8-6-4-2-10105-5
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⁡(x+x2+1)=0\log{\left(x + \sqrt{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 log(x + sqrt(1 + x^2)).
log⁡(02+1)\log{\left(\sqrt{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
xx2+1+1x+x2+1=0\frac{\frac{x}{\sqrt{x^{2} + 1}} + 1}{x + \sqrt{x^{2} + 1}} = 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
−x2x2+1−1x2+1+(xx2+1+1)2x+x2+1x+x2+1=0- \frac{\frac{\frac{x^{2}}{x^{2} + 1} - 1}{\sqrt{x^{2} + 1}} + \frac{\left(\frac{x}{\sqrt{x^{2} + 1}} + 1\right)^{2}}{x + \sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}} = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0

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
(−∞,0]\left(-\infty, 0\right]
Convex at the intervals
[0,∞)\left[0, \infty\right)
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞log⁡(x+x2+1)=−∞\lim_{x \to -\infty} \log{\left(x + \sqrt{x^{2} + 1} \right)} = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞log⁡(x+x2+1)=∞\lim_{x \to \infty} \log{\left(x + \sqrt{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 log(x + sqrt(1 + x^2)), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(log⁡(x+x2+1)x)=0\lim_{x \to -\infty}\left(\frac{\log{\left(x + \sqrt{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⁡(x+x2+1)x)=0\lim_{x \to \infty}\left(\frac{\log{\left(x + \sqrt{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⁡(x+x2+1)=log⁡(−x+x2+1)\log{\left(x + \sqrt{x^{2} + 1} \right)} = \log{\left(- x + \sqrt{x^{2} + 1} \right)}
- No
log⁡(x+x2+1)=−log⁡(−x+x2+1)\log{\left(x + \sqrt{x^{2} + 1} \right)} = - \log{\left(- x + \sqrt{x^{2} + 1} \right)}
- No
so, the function
not is
neither even, nor odd
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