Find the Range of Functions

The given function has a constant value 3 and therefore the range is the set:

\[ \{3\} \]

Assuming that the domain of the given function is the set of all real numbers \( \mathbb{R} \), so that the variable \( x \) takes all values in the interval \( (-\infty, +\infty) \).

If \( x \) takes all values in the interval \( (-\infty, +\infty) \), then \( 4x + 5 \) takes all values in the interval \( (-\infty, +\infty) \). Hence, the range of the given function is given by the interval:

\[ (-\infty, +\infty) \]

Assuming that the domain of the given function is \( \mathbb{R} \), meaning that \( x \) takes all values in the interval \( (-\infty, +\infty) \), which implies that \( x^2 \) is either zero or positive. Hence, we can write the inequality:

\[ x^2 \geq 0 \]

Add 5 to both sides of the inequality to obtain:

\[ x^2 + 5 \geq 5 \quad \text{or} \quad f(x) \geq 5 \]

The range of \( f(x) = x^2 + 5 \) is given by the interval:

\[ [5, +\infty) \]

We first write the given quadratic function in vertex form by completing the square:

\[ f(x) = -2x^2 + 4x - 7 = -2(x^2 - 2x) - 7 = -2((x - 1)^2 - 1) - 7 = -2(x - 1)^2 - 5 \]

The domain of the given function is \( \mathbb{R} \) with \( x \) taking any value in the interval \( (-\infty, +\infty) \), hence \( (x - 1)^2 \) is either zero or positive. We start by writing the inequality:

\[ (x - 1)^2 \geq 0 \]

Multiply both sides of the inequality by -2 and reverse the symbol of inequality to obtain:

\[ -2(x - 1)^2 \leq 0 \]

Add \(-5\) to both sides of the inequality to obtain:

\[ -2(x - 1)^2 - 5 \leq -5 \quad \text{or} \quad f(x) \leq -5 \]

Hence, the range of function \( f \) is given by the interval:

\[ (-\infty, -5] \]

The domain of the given function is \( \mathbb{R} \), and therefore \( x \) takes all values in the interval \( (-\infty, +\infty) \). The exponent \( 2x + 5 \) takes all values in \( (-\infty, +\infty) \). Since the basic exponential function is always positive, we can write:

\[ e^{2x + 5} > 0 \]

Multiply both sides of the inequality by -3 and reverse the symbol of inequality to obtain:

\[ -3e^{2x + 5} < 0 \]

Add 2 to both sides of the inequality to obtain:

\[ -3e^{2x + 5} + 2 < 2 \quad \text{or} \quad f(x) < 2 \]

The range of the given function is given by the interval:

\[ (-\infty, 2) \]

The domain of the given function is \( \mathbb{R} \). We start by writing the inequality:

\[ x^2 \geq 0 \]

Add 5 to both sides of the inequality:

\[ x^2 + 5 \geq 5 \]

Since the basic exponential function is an increasing function, we can apply it to write:

\[ e^{x^2 + 5} \geq e^5 \]

Multiply both sides of the inequality by -3 (reversing the inequality sign) and add 2 to both sides to obtain:

\[ -3e^{x^2 + 5} + 2 \leq -3e^5 + 2 \]

Note that the left-hand side of the inequality is equal to \( f(x) \). Hence:

\[ f(x) \leq -3e^5 + 2 \]

This means that the range of function \( f \) is given by the interval:

\[ (-\infty, -3e^5 + 2] \]

For this rational function, a direct algebraic method is not immediately obvious. Let us first find its inverse; the domain of its inverse will give the range of \( f \).

We first prove that \( f \) is a one-to-one function. For a function to be one-to-one, we need to show that if \( f(a) = f(b) \), then \( a = b \):

\[ \frac{a - 1}{a + 2} = \frac{b - 1}{b + 2} \]

Cross multiply, expand, and simplify:

\[ (a - 1)(b + 2) = (b - 1)(a + 2) \] \[ ab + 2a - b - 2 = ab + 2b - a - 2 \] \[ 3a = 3b \implies a = b \]

This proves that \( f \) is a one-to-one function. Now let us find the inverse of \( f \):

\[ y = \frac{x - 1}{x + 2} \]

Solve for \( x \):

\[ y(x + 2) = x - 1 \implies xy + 2y = x - 1 \implies x - xy = 2y + 1 \implies x(1 - y) = 2y + 1 \implies x = \frac{2y + 1}{1 - y} \]

Interchanging \( x \) and \( y \) gives the inverse function:

\[ f^{-1}(x) = \frac{2x + 1}{1 - x} \]

The range of \( f \) is given by the domain of \( f^{-1} \), which excludes \( x = 1 \). Therefore, the range is given by the interval:

\[ (-\infty, 1) \cup (1, +\infty) \]

The range of \( \ln(x) \) is given by the interval \( (-\infty, +\infty) \). Since the graph of \( \ln(x + 3) \) is the graph of \( \ln(x) \) shifted 3 units to the left, its range is also \( (-\infty, +\infty) \).

The graph of \( -3\ln(x + 3) \) is that of \( \ln(x + 3) \) reflected across the x-axis and vertically stretched by a factor of 3; its range remains \( (-\infty, +\infty) \).

Finally, shifting the graph 2 units down to obtain \( f(x) \) keeps the range unchanged. Thus, the range is:

\[ (-\infty, +\infty) \]

\( x^2 \) is a quantity that may be zero or positive, therefore we can write:

\[ x^2 \geq 0 \]

Add 4 to both sides of the inequality to obtain:

\[ x^2 + 4 \geq 4 \]

Divide both sides of the inequality by the positive quantity \( 4(x^2 + 4) \) (or take reciprocals, reversing the inequality) to obtain:

\[ \frac{1}{x^2 + 4} \leq \frac{1}{4} \]

Since \( \frac{1}{x^2 + 4} \) is always positive and never zero (though it gets very close to zero as \( x \) increases or decreases indefinitely), the range of \( f \) is given by the interval:

\[ \left(0, \frac{1}{4}\right] \]

Note that \( x^2 + 4x + 4 = (x + 2)^2 \). Therefore:

\[ f(x) = \sqrt{x^2 + 4x + 4} - 6 = \sqrt{(x + 2)^2} - 6 = |x + 2| - 6 \]

The range of \( |x + 2| \) is given by the interval \( [0, +\infty) \). The graph of \( f \) is that of \( |x + 2| \) shifted down by 6 units, and therefore the range of \( f \) is given by the interval:

\[ [-6, +\infty) \]

From basic trigonometry, we know that the range of values of the sine function is \( [-1, 1] \). Hence:

\[ -1 \leq \sin(3x - \pi) \leq 1 \]

Multiply all terms by -2 (reversing the inequalities) and add 1.5 to all terms:

\[ 2 + 1.5 \geq -2\sin(3x - \pi) + 1.5 \geq -2 + 1.5 \]

Simplify and rewrite as:

\[ -0.5 \leq f(x) \leq 3.5 \]

This gives the range of \( f \) as the interval:

\[ [-0.5, 3.5] \]