Answer:
- Domain: (-∞, ∞)
- Range: [0, ∞)
- Continuity: Function is continuous on its domain (-∞, ∞).
- Minimum stationary point (turning point) at (0, 0).
- Increasing function: (0, ∞)
- Decreasing function: (-∞, 0)
- Symmetry: Even (symmetry about the y-axis).
Step-by-step explanation:
Given function:
[tex]y=3x^2[/tex]
Domain
The domain is the set of all possible input values (x-values).
The domain of the given function is unrestricted.
Therefore, the domain is (-∞, ∞).
Range
The range is the set of all possible output values (y-values).
As x² ≥ 0, 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 (-∞, ∞).
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 & = 3x^2\\\implies \dfrac{\text{d}y}{\text{d}x} & = 6x\\\\\dfrac{\text{d}y}{\text{d}x} & = 0\\\implies 6x & = 0\\x & = 0\end{aligned}[/tex]
[tex]\textsf{When} \:x = 0 \implies y=3(0)^2=0[/tex]
Therefore, the stationary point of the given function is (0, 0).
To determine if a stationary point is minimum or maximum, differentiate the function again and substitute the x-value of the stationary point:
[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x} & = 6x\\\implies \dfrac{\text{d}^2y}{\text{d}x^2} & = 6 \end{aligned}[/tex]
As the second derivative is positive regardless of the x-value of the stationary point, the stationary point is a minimum.
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 6x & > 0 \\ x & > 0\end{aligned}[/tex]
Decreasing
[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x} & < 0 \\ \implies 6x & < 0 \\ x & < 0\end{aligned}[/tex]
Therefore:
- The function is increasing in the interval (0, ∞).
- The function is decreasing in the interval (-∞, 0).
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 x² ≥ 0 and there is symmetry about the y-axis,
[tex]\textsf{then $f(x)=f-x)$ for all $x$,}[/tex]
so the function is even.
