Mister Exam
Graphing y = 2^(x+1)+log(3)x+1
The solution
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x + 1 f(x) = 2 + log(3)*x + 1
$$f{\left(x \right)} = \left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1$$
f = 2^(x + 1) + x*log(3) + 1
The graph of the function
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:
$$\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1 = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \frac{1}{\log{\left(3 \right)}} - \frac{W\left(\frac{2 \log{\left(2 \right)}}{2^{\frac{1}{\log{\left(3 \right)}}} \log{\left(3 \right)}}\right)}{\log{\left(2 \right)}}$$
Numerical solution
$$x_{1} = -1.53740316130828$$
so we need to solve the equation:
$$\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1 = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = - \frac{1}{\log{\left(3 \right)}} - \frac{W\left(\frac{2 \log{\left(2 \right)}}{2^{\frac{1}{\log{\left(3 \right)}}} \log{\left(3 \right)}}\right)}{\log{\left(2 \right)}}$$
Numerical solution
$$x_{1} = -1.53740316130828$$
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
$$2^{x + 1} \log{\left(2 \right)} + \log{\left(3 \right)} = 0$$
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
$$2^{x + 1} \log{\left(2 \right)} + \log{\left(3 \right)} = 0$$
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
$$2 \cdot 2^{x} \log{\left(2 \right)}^{2} = 0$$
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
$$2 \cdot 2^{x} \log{\left(2 \right)}^{2} = 0$$
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
$$\lim_{x \to -\infty}\left(\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1\right) = \infty$$
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 2^(x + 1) + log(3)*x + 1, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1}{x}\right) = \log{\left(3 \right)}$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \log{\left(3 \right)}$$
$$\lim_{x \to \infty}\left(\frac{\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1}{x}\right) = \log{\left(3 \right)}$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \log{\left(3 \right)}$$
$$\lim_{x \to \infty}\left(\frac{\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1}{x}\right) = \infty$$
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:
$$\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1 = 2^{1 - x} - x \log{\left(3 \right)} + 1$$
- No
$$\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1 = - 2^{1 - x} + x \log{\left(3 \right)} - 1$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1 = 2^{1 - x} - x \log{\left(3 \right)} + 1$$
- No
$$\left(2^{x + 1} + x \log{\left(3 \right)}\right) + 1 = - 2^{1 - x} + x \log{\left(3 \right)} - 1$$
- No
so, the function
not is
neither even, nor odd