Mister Exam
Graphing y = sqrt(2*x+3)+sqrt(x+1)
The solution
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_________ _______ f(x) = \/ 2*x + 3 + \/ x + 1
$$f{\left(x \right)} = \sqrt{x + 1} + \sqrt{2 x + 3}$$
f = sqrt(x + 1) + sqrt(2*x + 3)
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:
$$\sqrt{x + 1} + \sqrt{2 x + 3} = 0$$
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:
$$\sqrt{x + 1} + \sqrt{2 x + 3} = 0$$
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
$$\frac{1}{\sqrt{2 x + 3}} + \frac{1}{2 \sqrt{x + 1}} = 0$$
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
$$\frac{1}{\sqrt{2 x + 3}} + \frac{1}{2 \sqrt{x + 1}} = 0$$
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
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(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
$$- (\frac{1}{\left(2 x + 3\right)^{\frac{3}{2}}} + \frac{1}{4 \left(x + 1\right)^{\frac{3}{2}}}) = 0$$
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
$$- (\frac{1}{\left(2 x + 3\right)^{\frac{3}{2}}} + \frac{1}{4 \left(x + 1\right)^{\frac{3}{2}}}) = 0$$
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
$$\lim_{x \to -\infty}\left(\sqrt{x + 1} + \sqrt{2 x + 3}\right) = \infty \operatorname{sign}{\left(i + \sqrt{2} i \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \infty \operatorname{sign}{\left(i + \sqrt{2} i \right)}$$
$$\lim_{x \to \infty}\left(\sqrt{x + 1} + \sqrt{2 x + 3}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
$$\lim_{x \to -\infty}\left(\sqrt{x + 1} + \sqrt{2 x + 3}\right) = \infty \operatorname{sign}{\left(i + \sqrt{2} i \right)}$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \infty \operatorname{sign}{\left(i + \sqrt{2} i \right)}$$
$$\lim_{x \to \infty}\left(\sqrt{x + 1} + \sqrt{2 x + 3}\right) = \infty$$
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(2*x + 3) + sqrt(x + 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x + 1} + \sqrt{2 x + 3}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{x + 1} + \sqrt{2 x + 3}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x + 1} + \sqrt{2 x + 3}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{x + 1} + \sqrt{2 x + 3}}{x}\right) = 0$$
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:
$$\sqrt{x + 1} + \sqrt{2 x + 3} = \sqrt{1 - x} + \sqrt{3 - 2 x}$$
- No
$$\sqrt{x + 1} + \sqrt{2 x + 3} = - \sqrt{1 - x} - \sqrt{3 - 2 x}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\sqrt{x + 1} + \sqrt{2 x + 3} = \sqrt{1 - x} + \sqrt{3 - 2 x}$$
- No
$$\sqrt{x + 1} + \sqrt{2 x + 3} = - \sqrt{1 - x} - \sqrt{3 - 2 x}$$
- No
so, the function
not is
neither even, nor odd