Mister Exam

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  • How to use it?

  • Graphing y =:
  • x³-6x
  • x^3-3x^2+1
  • (x^2)/(x-1)
  • x^2+5
  • Identical expressions

  • (e^x-e^(-x))/ two
  • (e to the power of x minus e to the power of ( minus x)) divide by 2
  • (e to the power of x minus e to the power of ( minus x)) divide by two
  • (ex-e(-x))/2
  • ex-e-x/2
  • e^x-e^-x/2
  • (e^x-e^(-x)) divide by 2
  • Similar expressions

  • (e^x-e^(x))/2
  • (e^x+e^(-x))/2

Graphing y = (e^x-e^(-x))/2

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
        x    -x
       E  - E  
f(x) = --------
          2    
$$f{\left(x \right)} = \frac{e^{x} - e^{- x}}{2}$$
f = (E^x - E^(-x))/2
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:
$$\frac{e^{x} - e^{- x}}{2} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 0$$
Numerical solution
$$x_{1} = 0$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (E^x - E^(-x))/2.
$$\frac{- e^{- 0} + e^{0}}{2}$$
The result:
$$f{\left(0 \right)} = 0$$
The point:
(0, 0)
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
$$\frac{e^{x}}{2} + \frac{e^{- x}}{2} = 0$$
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
$$\frac{e^{x} - e^{- x}}{2} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$

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
$$\left[0, \infty\right)$$
Convex at the intervals
$$\left(-\infty, 0\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{x} - e^{- x}}{2}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{e^{x} - e^{- x}}{2}\right) = \infty$$
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 (E^x - E^(-x))/2, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{e^{x} - e^{- x}}{2 x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{e^{x} - e^{- x}}{2 x}\right) = \infty$$
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:
$$\frac{e^{x} - e^{- x}}{2} = - \frac{e^{x}}{2} + \frac{e^{- x}}{2}$$
- No
$$\frac{e^{x} - e^{- x}}{2} = \frac{e^{x}}{2} - \frac{e^{- x}}{2}$$
- No
so, the function
not is
neither even, nor odd