Mister Exam
Other calculators
- Graph of implicit function
- Derivative Step by Step
- Integral Step by Step
- Graph of parametric function
- Graph in Polar Coordinates
- Surface defined parametrically
- Surface defined by equation
- Curve in 3D space defined parametrically
- Area between lines
- The intersection points of functions
-
How to use it?
- Graphing y =:
- -x^2(x+4)^2
- ((x^2+x-6)(x^2-2x-3))/(x^2-9)
- x^2-x+2
- -x^2+6x-8
- Derivative of:
-
e^(2x+1)
- Integral of d{x}:
-
e^(2x+1)
-
Identical expressions
- e^(2x+ one)
- e to the power of (2x plus 1)
- e to the power of (2x plus one)
- e(2x+1)
- e2x+1
- e^2x+1
-
Similar expressions
- e^(2x-1)
- -(2x+1)*e^2(x+1)
Graphing y = e^(2x+1)
The solution
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:
$$e^{2 x + 1} = 0$$
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:
$$e^{2 x + 1} = 0$$
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
$$2 e^{2 x + 1} = 0$$
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
$$2 e^{2 x + 1} = 0$$
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
$$4 e^{2 x + 1} = 0$$
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
$$4 e^{2 x + 1} = 0$$
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
$$\lim_{x \to -\infty} e^{2 x + 1} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} e^{2 x + 1} = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty} e^{2 x + 1} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{x \to \infty} e^{2 x + 1} = \infty$$
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 E^(2*x + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{2 x + 1}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{2 x + 1}}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\frac{e^{2 x + 1}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{e^{2 x + 1}}{x}\right) = \infty$$
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:
$$e^{2 x + 1} = e^{1 - 2 x}$$
- No
$$e^{2 x + 1} = - e^{1 - 2 x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$e^{2 x + 1} = e^{1 - 2 x}$$
- No
$$e^{2 x + 1} = - e^{1 - 2 x}$$
- No
so, the function
not is
neither even, nor odd