Question

NO LINKS!! Please help me with these problems. Part 11a1


y = 3x^2​

Answer 1

Domain: (−∞,∞) , x

Range: [0,∞) , y ≥ 0

Continuity: y=3x^2 is a parabola opening upwards, the vertex is at (0,0) and the focus is (0,1/12)

Max's and Min's: the minimum and maximum x value goes to infinity while for the y value the minimum is 0 and goes up to infinity.

Intervals: both x and y start to increase from the origin

Symmetry: the axis of symmetry is x=0 so therefore there the parabola is symmetrical

Answer 2

• 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).

Explanation:

Given function:

`y=3x^2`

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
`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 (-∞, ∞).

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 & = 3x^2\\\implies \frac{\text{d}y}{\text{d}x} & = 6x\\\\\frac{\text{d}y}{\text{d}x} & = 0\\\implies 6x & = 0\\x & = 0\end{aligned}`

`\textsf{When} \:x = 0 \implies y=3(0)^2=0`

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:

`\begin{aligned}\frac{\text{d}y}{\text{d}x} & = 6x\\\implies \frac{\text{d}^2y}{\text{d}x^2} & = 6 \end{aligned}`

As the second derivative is positive regardless of the x-value of the stationary point, the stationary point is a minimum.

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 6x & > 0 \\ x & > 0\end{aligned}`

Decreasing

`\begin{aligned}\frac{\text{d}y}{\text{d}x} & < 0 \\ \implies 6x & < 0 \\ x & < 0\end{aligned}`

Therefore:

• The function is increasing in the interval (0, ∞).
• The function is decreasing in the interval (-∞, 0).

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 x² ≥ 0 and there is symmetry about the y-axis,

`\textsf{then $f(x)=f-x)$ for all $x$,}`
so the function is even.