Question

NO LINKS!! Please help me with these problems. Part 10a1


y = √(x - 2)
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Answer 1

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.

Answer 2

• 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.

Explanation:

Given function:

`y=√(x-2)`

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.

`x-2\geq 0 \implies x\geq 2`

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
`c` in its domain:

`\text{$f(c)$ is de\:\!fined \quad and \quad $\displaystyle \lim_(x \to c) f(x) = f(c)$}`

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.

`\begin{aligned}y & = √(x-2)\\& = (x-2)^{(1)/(2)\\\implies \frac{\text{d}y}{\text{d}x} & = (1)/(2)(x-2)^{-(1)/(2)}\\ & = (1)/(2√(x-2))\\\\\frac{\text{d}y}{\text{d}x} & = 0\\\implies (1)/(2√(x-2)) &= 0\\(1)/(2) &\\eq 0\end{aligned}`

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

`\textsf{A function is \textbf{increasing} when the \underline{gradient is positive}} \implies \frac{\text{d}y}{\text{d}x} > 0`

`\textsf{A function is \textbf{decreasing} when the \underline{gradient is negative}} \implies \frac{\text{d}y}{\text{d}x} < 0`

Increasing

`\begin{aligned}\frac{\text{d}y}{\text{d}x} & > 0 \\ \implies (1)/(2√(x-2)) & > 0 \\ (1)/(√(x-2)) & > 0\\ \textsf{If $(1)/(a) > 0$ then $a > 0$}: \\ \implies √(x-2) & > 0\\ x-2 & > 0\\ x & > 2\end{aligned}`

Decreasing

`\begin{aligned}\frac{\text{d}y}{\text{d}x} & < 0 \\ \implies (1)/(2√(x-2)) & < 0 \\ (1)/(√(x-2)) & < 0\\ \textsf{If $(1)/(a) < 0$ then $a < 0$}: \\ \implies √(x-2) & < 0\\ \textsf{As $√(x)\geq 0$, no solution.}\end{aligned}`

Therefore:

• The function is increasing in the interval (2, ∞).

Symmetry

`\textsf{A function is \underline{even} when $f(x) = f(-x)$ for all $x$.}`

`\textsf{A function is \underline{odd} when $-f(x) = f(-x)$ for all $x$.}`

As there is no symmetry about the y-axis (even symmetry) and no origin symmetry (odd symmetry), the symmetry of the function is neither.