Answer:
- Domain: [2, ∞)
- Range: [0, ∞)
- Continuity: Function is continuous on its domain [2, ∞).
- Minimum point at (2, 0).
- Increasing function: (2, ∞)
- Symmetry: Neither as there is no symmetry about the y-axis and no origin symmetry.
Step-by-step explanation:
Given function:
[tex]y=\sqrt{x-2}[/tex]
Domain
The domain is the set of all possible input values (x-values).
As the square root of a negative number cannot be taken, the domain of the given function is restricted.
[tex]x-2\geq 0 \implies x\geq 2[/tex]
Therefore, the domain of the given function is [2, ∞).
Range
The range is the set of all possible output values (y-values).
As the square root of a negative number cannot be taken, the range of the given function is restricted.
Therefore, the range of the given function is [0, ∞).
Continuity
A function f(x) is continuous when, for every value [tex]c[/tex] in its domain:
[tex]\text{$f(c)$ is de\:\!fined \quad and \quad $\displaystyle \lim_{x \to c} f(x) = f(c)$}[/tex]
Therefore, the function is continuous on its domain [2, ∞).
Maximums and Minimums
Stationary points occur when the gradient of a graph is zero.
Therefore, to find the x-coordinate(s) of the stationary points of a function, differentiate the function, set it to zero and solve for x.
[tex]\begin{aligned}y & = \sqrt{x-2}\\& = (x-2)^{\frac{1}{2}\\\implies \dfrac{\text{d}y}{\text{d}x} & = \dfrac{1}{2}(x-2)^{-\frac{1}{2}}\\ & = \dfrac{1}{2\sqrt{x-2}}\\\\\dfrac{\text{d}y}{\text{d}x} & = 0\\\implies \dfrac{1}{2\sqrt{x-2}} &= 0\\\dfrac{1}{2} &\neq 0\end{aligned}[/tex]
Therefore, there are no stationary points.
As the domain is restricted and f(x) ≥ 0, the minimum point of the function is at point (2, 0).
Increasing/Decreasing Function
[tex]\textsf{A function is \textbf{increasing} when the \underline{gradient is positive}} \implies \dfrac{\text{d}y}{\text{d}x} > 0[/tex]
[tex]\textsf{A function is \textbf{decreasing} when the \underline{gradient is negative}} \implies \dfrac{\text{d}y}{\text{d}x} < 0[/tex]
Increasing
[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x} & > 0 \\ \implies \dfrac{1}{2\sqrt{x-2}} & > 0 \\ \dfrac{1}{\sqrt{x-2}} & > 0\\ \textsf{If $\frac{1}{a} > 0$ then $a > 0$}: \\ \implies \sqrt{x-2} & > 0\\ x-2 & > 0\\ x & > 2\end{aligned}[/tex]
Decreasing
[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x} & < 0 \\ \implies \dfrac{1}{2\sqrt{x-2}} & < 0 \\ \dfrac{1}{\sqrt{x-2}} & < 0\\ \textsf{If $\frac{1}{a} < 0$ then $a < 0$}: \\ \implies \sqrt{x-2} & < 0\\ \textsf{As $\sqrt{x}\geq 0$, no solution.}\end{aligned}[/tex]
Therefore:
- The function is increasing in the interval (2, ∞).
Symmetry
[tex]\textsf{A function is \underline{even} when $f(x) = f(-x)$ for all $x$.}[/tex]
[tex]\textsf{A function is \underline{odd} when $-f(x) = f(-x)$ for all $x$.}[/tex]
As there is no symmetry about the y-axis (even symmetry) and no origin symmetry (odd symmetry), the symmetry of the function is neither.
