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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = log(x)*\/  x  + x + 1 
$$f{\left(x \right)} = \sqrt{\left(x^{2} + x\right) + 1} \log{\left(x \right)}$$
f = sqrt(x^2 + x + 1)*log(x)
The graph of the function
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:
$$\sqrt{\left(x^{2} + x\right) + 1} \log{\left(x \right)} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 1$$
Numerical solution
$$x_{1} = 1$$
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(x^2 + x + 1).
$$\sqrt{0^{2} + 1} \log{\left(0 \right)}$$
The result:
$$f{\left(0 \right)} = \tilde{\infty}$$
sof doesn't intersect Y
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$\frac{\left(x + \frac{1}{2}\right) \log{\left(x \right)}}{\sqrt{\left(x^{2} + x\right) + 1}} + \frac{\sqrt{\left(x^{2} + x\right) + 1}}{x} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$- \frac{\left(\frac{\left(2 x + 1\right)^{2}}{x^{2} + x + 1} - 4\right) \log{\left(x \right)}}{4 \sqrt{x^{2} + x + 1}} + \frac{2 x + 1}{x \sqrt{x^{2} + x + 1}} - \frac{\sqrt{x^{2} + x + 1}}{x^{2}} = 0$$
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
$$\lim_{x \to -\infty}\left(\sqrt{\left(x^{2} + x\right) + 1} \log{\left(x \right)}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\sqrt{\left(x^{2} + x\right) + 1} \log{\left(x \right)}\right) = \infty$$
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(x^2 + x + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sqrt{\left(x^{2} + x\right) + 1} \log{\left(x \right)}}{x}\right) = -\infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\sqrt{\left(x^{2} + x\right) + 1} \log{\left(x \right)}}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
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:
$$\sqrt{\left(x^{2} + x\right) + 1} \log{\left(x \right)} = \sqrt{x^{2} - x + 1} \log{\left(- x \right)}$$
- No
$$\sqrt{\left(x^{2} + x\right) + 1} \log{\left(x \right)} = - \sqrt{x^{2} - x + 1} \log{\left(- x \right)}$$
- No
so, the function
not is
neither even, nor odd