Mister Exam

Graphing y = lnx+1

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = log(x) + 1
f(x)=log(x)+1f{\left(x \right)} = \log{\left(x \right)} + 1
f = log(x) + 1
The graph of the function
2.00.00.20.40.60.81.01.21.41.61.8-1010
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)+1=0\log{\left(x \right)} + 1 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=e1x_{1} = e^{-1}
Numerical solution
x1=0.367879441171442x_{1} = 0.367879441171442
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log(x) + 1.
log(0)+1\log{\left(0 \right)} + 1
The result:
f(0)=~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
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
1x=0\frac{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
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
1x2=0- \frac{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
limx(log(x)+1)=\lim_{x \to -\infty}\left(\log{\left(x \right)} + 1\right) = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
limx(log(x)+1)=\lim_{x \to \infty}\left(\log{\left(x \right)} + 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) + 1, divided by x at x->+oo and x ->-oo
limx(log(x)+1x)=0\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} + 1}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
limx(log(x)+1x)=0\lim_{x \to \infty}\left(\frac{\log{\left(x \right)} + 1}{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)+1=log(x)+1\log{\left(x \right)} + 1 = \log{\left(- x \right)} + 1
- No
log(x)+1=log(x)1\log{\left(x \right)} + 1 = - \log{\left(- x \right)} - 1
- No
so, the function
not is
neither even, nor odd