Mister Exam

Graphing y = 2^(5-x)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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        5 - x
f(x) = 2     
$$f{\left(x \right)} = 2^{5 - x}$$
f = 2^(5 - x)
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:
$$2^{5 - x} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 2^(5 - x).
$$2^{5 - 0}$$
The result:
$$f{\left(0 \right)} = 32$$
The point:
(0, 32)
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
$$- 2^{5 - x} \log{\left(2 \right)} = 0$$
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
$$32 \cdot 2^{- x} \log{\left(2 \right)}^{2} = 0$$
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
$$\lim_{x \to -\infty} 2^{5 - x} = \infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} 2^{5 - x} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 2^(5 - x), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{2^{5 - x}}{x}\right) = -\infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{2^{5 - x}}{x}\right) = 0$$
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:
$$2^{5 - x} = 2^{x + 5}$$
- No
$$2^{5 - x} = - 2^{x + 5}$$
- No
so, the function
not is
neither even, nor odd