Mister Exam
Graphing y = y=2^(2*sqrt(x))
The solution
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2*\/ x
f(x) = 2
$$f{\left(x \right)} = 2^{2 \sqrt{x}}$$
f = 2^(2*sqrt(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:
$$2^{2 \sqrt{x}} = 0$$
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:
$$2^{2 \sqrt{x}} = 0$$
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
$$\frac{2^{2 \sqrt{x}} \log{\left(2 \right)}}{\sqrt{x}} = 0$$
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
$$\frac{2^{2 \sqrt{x}} \log{\left(2 \right)}}{\sqrt{x}} = 0$$
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
$$2^{2 \sqrt{x}} \left(\frac{\log{\left(2 \right)}}{x} - \frac{1}{2 x^{\frac{3}{2}}}\right) \log{\left(2 \right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{1}{4 \log{\left(2 \right)}^{2}}$$
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
$$\left[\frac{1}{4 \log{\left(2 \right)}^{2}}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{1}{4 \log{\left(2 \right)}^{2}}\right]$$
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$2^{2 \sqrt{x}} \left(\frac{\log{\left(2 \right)}}{x} - \frac{1}{2 x^{\frac{3}{2}}}\right) \log{\left(2 \right)} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = \frac{1}{4 \log{\left(2 \right)}^{2}}$$
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
$$\left[\frac{1}{4 \log{\left(2 \right)}^{2}}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{1}{4 \log{\left(2 \right)}^{2}}\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
Limit on the left could not be calculated
$$\lim_{x \to -\infty} 2^{2 \sqrt{x}}$$
$$\lim_{x \to \infty} 2^{2 \sqrt{x}} = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Limit on the left could not be calculated
$$\lim_{x \to -\infty} 2^{2 \sqrt{x}}$$
$$\lim_{x \to \infty} 2^{2 \sqrt{x}} = \infty$$
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^(2*sqrt(x)), divided by x at x->+oo and x ->-oo
Limit on the left could not be calculated
$$\lim_{x \to -\infty}\left(\frac{2^{2 \sqrt{x}}}{x}\right)$$
$$\lim_{x \to \infty}\left(\frac{2^{2 \sqrt{x}}}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
Limit on the left could not be calculated
$$\lim_{x \to -\infty}\left(\frac{2^{2 \sqrt{x}}}{x}\right)$$
$$\lim_{x \to \infty}\left(\frac{2^{2 \sqrt{x}}}{x}\right) = \infty$$
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:
$$2^{2 \sqrt{x}} = 2^{2 \sqrt{- x}}$$
- No
$$2^{2 \sqrt{x}} = - 2^{2 \sqrt{- x}}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$2^{2 \sqrt{x}} = 2^{2 \sqrt{- x}}$$
- No
$$2^{2 \sqrt{x}} = - 2^{2 \sqrt{- x}}$$
- No
so, the function
not is
neither even, nor odd