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+6x-9
- sin(x)
- y=x²-11x-2|x-5|+30
- y=2x^3+3x^2+8
- Derivative of:
-
e^(z-1)
-
Identical expressions
- e^(z- one)
- e to the power of (z minus 1)
- e to the power of (z minus one)
- e(z-1)
- ez-1
- e^z-1
-
Similar expressions
- e^(z+1)
Graphing y = e^(z-1)
The solution
The graph of the function
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis Z at f = 0
so we need to solve the equation:
$$e^{z - 1} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis Z
so we need to solve the equation:
$$e^{z - 1} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis Z
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d z} f{\left(z \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d z} f{\left(z \right)} = $$
the first derivative
$$e^{z - 1} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d z} f{\left(z \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d z} f{\left(z \right)} = $$
the first derivative
$$e^{z - 1} = 0$$
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
$$\frac{d^{2}}{d z^{2}} f{\left(z \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d z^{2}} f{\left(z \right)} = $$
the second derivative
$$e^{z - 1} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d z^{2}} f{\left(z \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d z^{2}} f{\left(z \right)} = $$
the second derivative
$$e^{z - 1} = 0$$
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 z->+oo and z->-oo
$$\lim_{z \to -\infty} e^{z - 1} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{z \to \infty} e^{z - 1} = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{z \to -\infty} e^{z - 1} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 0$$
$$\lim_{z \to \infty} e^{z - 1} = \infty$$
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^(z - 1), divided by z at z->+oo and z ->-oo
$$\lim_{z \to -\infty}\left(\frac{e^{z - 1}}{z}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{z \to \infty}\left(\frac{e^{z - 1}}{z}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
$$\lim_{z \to -\infty}\left(\frac{e^{z - 1}}{z}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{z \to \infty}\left(\frac{e^{z - 1}}{z}\right) = \infty$$
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(-z) и f = -f(-z).
So, check:
$$e^{z - 1} = e^{- z - 1}$$
- No
$$e^{z - 1} = - e^{- z - 1}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$e^{z - 1} = e^{- z - 1}$$
- No
$$e^{z - 1} = - e^{- z - 1}$$
- No
so, the function
not is
neither even, nor odd