Find the range of real valued mathematical functions using different techniques.

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 \( (-\infty , +\infty) \) and 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 means that \( x^2 \) is either zero or positive. Hence we can write the following inequality

\( x^2 \geq 0 \)

Add 5 to both sides of the inequality to obtain the inequality

\( x^2 + 5 \geq 5 \) or \( f(x) \geq 5 \)

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 change the symbol of inequality to obtain

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

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

\( -2 (x - 1)^2 - 5 \leq -5 \) or \( f(x) \leq -5 \) and 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 the interval \( (-\infty , +\infty) \), we can therefore write the following inequality (basic exponential function always positive)

\( e^{2x + 5} > 0 \)

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

\( -3e^{2x + 5} < 0 \)

Add 2 to both sides of the inequality to obtain

\( -3e^{2x + 5} + 2 \lt 2 \) or \( f(x) \lt 2 \)

The range of the given function is given by the interval \( (-\infty , 2) \).

The domain of the given function is \( \mathbb{R} \), we can start by writing the inequality

\( x^2 \geq 0 \)

Add 5 to both sides of the inequality

\( x^2 + 5 \geq 5 \)

The basic exponential function being an increasing function, we can use the above to write the inequality

\( e^{x^2 + 5} \geq e^5 \)

Multiply both sides of the inequality by -3 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

\( f(x) \lt -3e^5 + 2 \) which means that the range of function \( f \) is given by the interval: \( (-\infty , -3e^5 + 2) \)

For this rational function, a direct algebraic method similar to those above is not obvious. Let us first find its inverse, the domain of its inverse which give the range of \( f \).

We first prove that \( f \) is a one to one function and then find its inverse. For a function to be a one to one, we need to show that

If \( f(a) = f(b) \) then \( a = b \).

\( \dfrac{a - 1}{a + 2} = \dfrac{b - 1}{b + 2} \)

Cross multiply, expand and simplify

\( (a - 1)(b + 2) = (b - 1)(a + 2) \)

\( ab + 2a - b - 2 = ab + 2b - a - 2 \)

\( 3a = 3b \), which finally gives \( a = b \) and proves that \( f \) is a one to one function.

Let us find the inverse of \( f \)

\( y = \dfrac{x - 1}{x + 2} \)

Solve for \( x \)

\( x = \dfrac{2y + 1}{1 - y} \)

change \( y \) into \( x \) and \( x \) into \( y \) and write the inverse function

\( f^{-1}(x) = y = \dfrac{2x + 1}{1 - x} \)

The range of \( f \) is given by the domain of \( f^{-1} \) and is therefore 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, the range of \( \ln(x + 3) \) is also given by the interval \( (-\infty , +\infty) \).

The graph of \( -3\ln(x + 3) \) is that of \( \ln(x + 3) \) reflected on the x-axis because of the minus sign and expanded vertically by 3 and therefore the range is still given by the interval \( (-\infty , +\infty) \).

The graph of \( f \) is that of \( -3\ln(x +3) \) shifted 2 units up and therefore the range is also given by the interval \( (-\infty , +\infty) \).

\( x^2 \) is a quantity which may be zero or positive and 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 side of the inequality \( x^2 + 4 \geq 4 \) by the positive quantity \( 4(x^2 + 4) \) to obtain

\( \dfrac{1}{4} \geq \dfrac{1}{x^2 + 4} \)

The right side of the inequality is equal to \( f(x) \). Hence the above inequality may be written as

\( f(x) \leq \dfrac{1}{4} \)

Because \( \dfrac{1}{x^2 + 4} \) is always positive and never zero but may be very close to zero as \( x \) increases or decreases indefinitely , the range of \( f \) is given by the interval \( (0; \dfrac{1}{4}) \)

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 is given by the interval \( [-6 , +\infty) \)

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

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

Multiply all terms by -2; change symbols of 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

\( -0.5 \leq f(x) \leq 3.5 \) which gives the range of \( f \) as the interval \( [-0.5 , 3.5] \).