Mister Exam
Graphing y = (sqrtx+2)²-1
The solution
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2
/ ___ \
f(x) = \\/ x + 2/ - 1
$$f{\left(x \right)} = \left(\sqrt{x} + 2\right)^{2} - 1$$
f = (sqrt(x) + 2)^2 - 1
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:
$$\left(\sqrt{x} + 2\right)^{2} - 1 = 0$$
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:
$$\left(\sqrt{x} + 2\right)^{2} - 1 = 0$$
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
$$\frac{\sqrt{x} + 2}{\sqrt{x}} = 0$$
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
$$\frac{\sqrt{x} + 2}{\sqrt{x}} = 0$$
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
$$\frac{\frac{1}{x} - \frac{\sqrt{x} + 2}{x^{\frac{3}{2}}}}{2} = 0$$
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
$$\frac{\frac{1}{x} - \frac{\sqrt{x} + 2}{x^{\frac{3}{2}}}}{2} = 0$$
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
$$\lim_{x \to -\infty}\left(\left(\sqrt{x} + 2\right)^{2} - 1\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\left(\sqrt{x} + 2\right)^{2} - 1\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\left(\sqrt{x} + 2\right)^{2} - 1\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\left(\sqrt{x} + 2\right)^{2} - 1\right) = \infty$$
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 (sqrt(x) + 2)^2 - 1, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(\sqrt{x} + 2\right)^{2} - 1}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x} + 2\right)^{2} - 1}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
$$\lim_{x \to -\infty}\left(\frac{\left(\sqrt{x} + 2\right)^{2} - 1}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x} + 2\right)^{2} - 1}{x}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
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:
$$\left(\sqrt{x} + 2\right)^{2} - 1 = \left(\sqrt{- x} + 2\right)^{2} - 1$$
- No
$$\left(\sqrt{x} + 2\right)^{2} - 1 = 1 - \left(\sqrt{- x} + 2\right)^{2}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\left(\sqrt{x} + 2\right)^{2} - 1 = \left(\sqrt{- x} + 2\right)^{2} - 1$$
- No
$$\left(\sqrt{x} + 2\right)^{2} - 1 = 1 - \left(\sqrt{- x} + 2\right)^{2}$$
- No
so, the function
not is
neither even, nor odd