Mister Exam
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-
How to use it?
- Graphing y =:
- x^3+6*x^2+9*x+4
- x^3+4x
-
|x^2+x-2|
- x(2-x)^2
-
Identical expressions
- log one / three *(x+ three)-1
- logarithm of 1 divide by 3 multiply by (x plus 3) minus 1
- logarithm of one divide by three multiply by (x plus three) minus 1
- log1/3(x+3)-1
- log1/3x+3-1
- log1 divide by 3*(x+3)-1
-
Similar expressions
- log(1)/3*(x+3)-1
- log1/3*(x+3)+1
- log1/3*(x-3)-1
Graphing y = log1/3*(x+3)-1
The solution
You have entered
[src]
log(1)*(x + 3)
f(x) = -------------- - 1
3
$$f{\left(x \right)} = \frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1$$
f = log(1)*(x + 3)/3 - 1*1
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:
$$\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1 = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
so we need to solve the equation:
$$\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1 = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
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
$$\frac{\log{\left(1 \right)}}{3} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(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
$$\frac{\log{\left(1 \right)}}{3} = 0$$
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
$$0 = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$0 = 0$$
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
$$\lim_{x \to -\infty}\left(\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1\right) = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -1$$
$$\lim_{x \to \infty}\left(\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1\right) = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = -1$$
$$\lim_{x \to -\infty}\left(\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1\right) = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = -1$$
$$\lim_{x \to \infty}\left(\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1\right) = -1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = -1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of log(1)*(x + 3)/3 - 1*1, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1}{x}\right) = 0$$
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:
$$\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1 = \frac{\left(- x + 3\right) \log{\left(1 \right)}}{3} - 1$$
- No
$$\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1 = - \frac{\left(- x + 3\right) \log{\left(1 \right)}}{3} + 1$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1 = \frac{\left(- x + 3\right) \log{\left(1 \right)}}{3} - 1$$
- No
$$\frac{\left(x + 3\right) \log{\left(1 \right)}}{3} - 1 = - \frac{\left(- x + 3\right) \log{\left(1 \right)}}{3} + 1$$
- No
so, the function
not is
neither even, nor odd