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Plotting a function graph/ 3*(log(3)(1-x/2))+1

Graphing y = 3*(log(3)(1-x/2))+1

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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                /    x\    
f(x) = 3*log(3)*|1 - -| + 1
                \    2/
f(x)=3(−x2+1)log⁡(3)+1f{\left(x \right)} = 3 \left(- \frac{x}{2} + 1\right) \log{\left(3 \right)} + 1 
f = 3*((-x/2 + 1)*log(3)) + 1
The graph of the function
02468-8-6-4-2-1010-5050
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:
3(−x2+1)log⁡(3)+1=03 \left(- \frac{x}{2} + 1\right) \log{\left(3 \right)} + 1 = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=23log⁡(3)+2x_{1} = \frac{2}{3 \log{\left(3 \right)}} + 2
Numerical solution
x1=2.60682615108456x_{1} = 2.60682615108456
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 3*(log(3)*(1 - x/2)) + 1.
1+3(−02+1)log⁡(3)1 + 3 \left(- \frac{0}{2} + 1\right) \log{\left(3 \right)}
The result:
f(0)=1+3log⁡(3)f{\left(0 \right)} = 1 + 3 \log{\left(3 \right)}
The point:
(0, 1 + 3*log(3))
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
−3log⁡(3)2=0- \frac{3 \log{\left(3 \right)}}{2} = 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
0=00 = 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→−∞(3(−x2+1)log⁡(3)+1)=∞\lim_{x \to -\infty}\left(3 \left(- \frac{x}{2} + 1\right) \log{\left(3 \right)} + 1\right) = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞(3(−x2+1)log⁡(3)+1)=−∞\lim_{x \to \infty}\left(3 \left(- \frac{x}{2} + 1\right) \log{\left(3 \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 3*(log(3)*(1 - x/2)) + 1, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(3(−x2+1)log⁡(3)+1x)=−3log⁡(3)2\lim_{x \to -\infty}\left(\frac{3 \left(- \frac{x}{2} + 1\right) \log{\left(3 \right)} + 1}{x}\right) = - \frac{3 \log{\left(3 \right)}}{2}
Let's take the limit
so,
inclined asymptote equation on the left:
y=−3xlog⁡(3)2y = - \frac{3 x \log{\left(3 \right)}}{2}
lim⁡x→∞(3(−x2+1)log⁡(3)+1x)=−3log⁡(3)2\lim_{x \to \infty}\left(\frac{3 \left(- \frac{x}{2} + 1\right) \log{\left(3 \right)} + 1}{x}\right) = - \frac{3 \log{\left(3 \right)}}{2}
Let's take the limit
so,
inclined asymptote equation on the right:
y=−3xlog⁡(3)2y = - \frac{3 x \log{\left(3 \right)}}{2}
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:
3(−x2+1)log⁡(3)+1=3(x2+1)log⁡(3)+13 \left(- \frac{x}{2} + 1\right) \log{\left(3 \right)} + 1 = 3 \left(\frac{x}{2} + 1\right) \log{\left(3 \right)} + 1
- No
3(−x2+1)log⁡(3)+1=−3(x2+1)log⁡(3)−13 \left(- \frac{x}{2} + 1\right) \log{\left(3 \right)} + 1 = - 3 \left(\frac{x}{2} + 1\right) \log{\left(3 \right)} - 1
- No
so, the function
not is
neither even, nor odd
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