Mister Exam
Graphing y = log3((9x-1)/(x+2))
The solution
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/9*x - 1\
log|-------|
\ x + 2 /
f(x) = ------------
log(3)
$$f{\left(x \right)} = \frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}}$$
f = log((9*x - 1)/(x + 2))/log(3)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -2$$
$$x_{1} = -2$$
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:
$$\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = \frac{3}{8}$$
Numerical solution
$$x_{1} = 0.375$$
so we need to solve the equation:
$$\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = \frac{3}{8}$$
Numerical solution
$$x_{1} = 0.375$$
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
$$\frac{\left(x + 2\right) \left(\frac{9}{x + 2} - \frac{9 x - 1}{\left(x + 2\right)^{2}}\right)}{\left(9 x - 1\right) \log{\left(3 \right)}} = 0$$
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
$$\frac{\left(x + 2\right) \left(\frac{9}{x + 2} - \frac{9 x - 1}{\left(x + 2\right)^{2}}\right)}{\left(9 x - 1\right) \log{\left(3 \right)}} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$- \frac{\left(9 - \frac{9 x - 1}{x + 2}\right) \left(\frac{9}{9 x - 1} + \frac{1}{x + 2}\right)}{\left(9 x - 1\right) \log{\left(3 \right)}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{17}{18}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$\lim_{x \to -2^-}\left(- \frac{\left(9 - \frac{9 x - 1}{x + 2}\right) \left(\frac{9}{9 x - 1} + \frac{1}{x + 2}\right)}{\left(9 x - 1\right) \log{\left(3 \right)}}\right) = \infty$$
$$\lim_{x \to -2^+}\left(- \frac{\left(9 - \frac{9 x - 1}{x + 2}\right) \left(\frac{9}{9 x - 1} + \frac{1}{x + 2}\right)}{\left(9 x - 1\right) \log{\left(3 \right)}}\right) = \infty$$
- limits are equal, then skip the corresponding point
Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
$$\left(-\infty, - \frac{17}{18}\right]$$
Convex at the intervals
$$\left[- \frac{17}{18}, \infty\right)$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$- \frac{\left(9 - \frac{9 x - 1}{x + 2}\right) \left(\frac{9}{9 x - 1} + \frac{1}{x + 2}\right)}{\left(9 x - 1\right) \log{\left(3 \right)}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{17}{18}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$\lim_{x \to -2^-}\left(- \frac{\left(9 - \frac{9 x - 1}{x + 2}\right) \left(\frac{9}{9 x - 1} + \frac{1}{x + 2}\right)}{\left(9 x - 1\right) \log{\left(3 \right)}}\right) = \infty$$
$$\lim_{x \to -2^+}\left(- \frac{\left(9 - \frac{9 x - 1}{x + 2}\right) \left(\frac{9}{9 x - 1} + \frac{1}{x + 2}\right)}{\left(9 x - 1\right) \log{\left(3 \right)}}\right) = \infty$$
- limits are equal, then skip the corresponding point
Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
$$\left(-\infty, - \frac{17}{18}\right]$$
Convex at the intervals
$$\left[- \frac{17}{18}, \infty\right)$$
Vertical asymptotes
Have:
$$x_{1} = -2$$
$$x_{1} = -2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}}\right) = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 2$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}}\right) = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 2$$
$$\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}}\right) = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 2$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}}\right) = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 2$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of log((9*x - 1)/(x + 2))/log(3), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{x \log{\left(3 \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{x \log{\left(3 \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{x \log{\left(3 \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{x \log{\left(3 \right)}}\right) = 0$$
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:
$$\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(\frac{- 9 x - 1}{2 - x} \right)}}{\log{\left(3 \right)}}$$
- No
$$\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}} = - \frac{\log{\left(\frac{- 9 x - 1}{2 - x} \right)}}{\log{\left(3 \right)}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(\frac{- 9 x - 1}{2 - x} \right)}}{\log{\left(3 \right)}}$$
- No
$$\frac{\log{\left(\frac{9 x - 1}{x + 2} \right)}}{\log{\left(3 \right)}} = - \frac{\log{\left(\frac{- 9 x - 1}{2 - x} \right)}}{\log{\left(3 \right)}}$$
- No
so, the function
not is
neither even, nor odd