Mister Exam

Other calculators

Graphing y = log3^(x+2)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
          x + 2   
f(x) = log     (3)
f(x)=log(3)x+2f{\left(x \right)} = \log{\left(3 \right)}^{x + 2}
f = log(3)^(x + 2)
The graph of the function
02468-8-6-4-2-10100.05.0
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:
log(3)x+2=0\log{\left(3 \right)}^{x + 2} = 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 log(3)^(x + 2).
log(3)2\log{\left(3 \right)}^{2}
The result:
f(0)=log(3)2f{\left(0 \right)} = \log{\left(3 \right)}^{2}
The point:
(0, log(3)^2)
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
log(3)x+2log(log(3))=0\log{\left(3 \right)}^{x + 2} \log{\left(\log{\left(3 \right)} \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
log(3)x+2log(log(3))2=0\log{\left(3 \right)}^{x + 2} \log{\left(\log{\left(3 \right)} \right)}^{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
limxlog(3)x+2=0\lim_{x \to -\infty} \log{\left(3 \right)}^{x + 2} = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
limxlog(3)x+2=\lim_{x \to \infty} \log{\left(3 \right)}^{x + 2} = \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(3)^(x + 2), divided by x at x->+oo and x ->-oo
limx(log(3)x+2x)=0\lim_{x \to -\infty}\left(\frac{\log{\left(3 \right)}^{x + 2}}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
limx(log(3)x+2x)=\lim_{x \to \infty}\left(\frac{\log{\left(3 \right)}^{x + 2}}{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:
log(3)x+2=log(3)2x\log{\left(3 \right)}^{x + 2} = \log{\left(3 \right)}^{2 - x}
- No
log(3)x+2=log(3)2x\log{\left(3 \right)}^{x + 2} = - \log{\left(3 \right)}^{2 - x}
- No
so, the function
not is
neither even, nor odd