The Definition Of Differentiable Also Defines An Error Term E(x,y). Find E(x,y) For The Function F(x,y)=8x^2

Answer:

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

Step-by-step 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:

\(\implies \sin \left(\dfrac{t}{2}\right)=1\)

\(\implies \dfrac{t}{2}=\dfrac{\pi}{2}+2\pi n\)

\(\implies t=\pi+4\pi n\)

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