Question

Hey, Can anyone assist me with a bunch of calculus questions, thank you in advance

Answer 1

1. (a) [-1, ∞)

(b) (-∞, -1) ∪ (1, ∞)

2. (a) (1, 3)

(b) (-∞, 1) ∪ (3, ∞)

3. (a) 9.6 m and 0.4 m

(b) 03:08 and 15:42

Explanation:

The domain of a function is the set of all possible input values (x-values).

The range of a function is the set of all possible output values (y-values).

Question 1

Part (a)

When x < 0, the function is f(x) = x².

Since the square of any non-zero real number is always positive, the range of the function f(x) for x < 0 is (0, ∞).

When x ≥ 0, the function is f(x) = sin(x).

The minimum value of the sine function is -1 and the maximum value of the sine function is 1. As the sine function is periodic, the function oscillates between these values. Therefore, the range of function f(x) for x ≥ 0 is [-1, 1].

The range of function f(x) is the union of the ranges of the two separate parts of the function. Therefore, the range of f(x) is [-1, ∞).

Part (b)

The domain of the function g(x) = ln(x² - 1) is the set of all real numbers x for which (x² - 1) is positive, since the natural logarithm function (ln) is only defined for positive input values.

Find the values of x:

`\implies x^2-1 > 0`

`\implies x^2 > 1`

`\implies x < -1, \;\;x > 1`

Therefore, the domain of function g(x) is (-∞, -1) ∪ (1, ∞).

Question 2

Part (a)

To determine the interval where f(x) < 0, we need to find the values of x for which the quadratic is less than zero.

First, set the function equal to zero and solve for x:

`\begin{aligned} x^2-4x+3&=0\\x^2-3x-x+3&=0\\x(x-3)-1(x-3)&=0\\(x-1)(x-3)&=0\\ \implies x&=1,\;3\end{aligned}`

Therefore, the function is equal to zero at x = 1 and x = 3 and so the parabola crosses the x-axis at x = 1 and x = 3.

As the leading coefficient of the quadratic is positive, the parabola opens upwards. Therefore, the values of x that make the function negative are between the zeros. So the interval where f(x) < 0 is 1 < x < 3 = (1, 3).

Part (b)
Since the square root of a negative number cannot be taken, and dividing a number by zero is undefined, function f(x) has to be positive and not equal to zero: f(x) > 0.

As the parabola opens upwards, the values of x that make the function positive are less than the zero at x = 1 and more than the zero at x = 3.

Therefore the domain of g(x) is (-∞, 1) ∪ (3, ∞).

Question 3

Part (a)

The range of a sine function is [-1, 1]. Therefore, to calculate the maximal and minimal possible water depths of the bay, substitute the maximum and minimum values of sin(t/2) into the equation:

`\textsf{Maximum}: \quad 5+4.6(1)=9.6\; \sf m`

`\textsf{Maximum}: \quad 5+4.6(-1)=0.4\; \sf m`

Part (b)

To find the times when the depth is maximal, set sin(t/2) to 1 and solve for t:

`\implies \sin \left((t)/(2)\right)=1`

`\implies (t)/(2)=(\pi)/(2)+2\pi n`

`\implies t=\pi+4\pi n`

Therefore, the values of t in the interval 0 ≤ t ≤ 24 are:

• 
  `t = \pi=3.14159265...\sf hours\;after\;mindnight`
• 
  `t=5 \pi = 15.7079632...\sf hours\;after\;mindnight`

Convert these values to times:

• 03:08 and 15:42