Evaluate Functions

Evaluate function \( f \) for \( x = -2 \) and \( x = 2 \), if possible, given that \( f \) is defined by:

\[ f(x) = \dfrac{-4}{x + 2} \]

Solution:

Function \( f \) given above has the domain:

\[ (-\infty, -2) \cup (-2, +\infty) \]

Since at \( x = -2 \) the denominator of \( f(x) \) is equal to 0 (which is outside the domain):

\[ f(-2) = \text{undefined} \]

To find \( f(2) \), substitute \( x \) by \( 2 \) in \( f(x) \):

\[ f(2) = \dfrac{-4}{2 + 2} = \dfrac{-4}{4} = -1 \]

Evaluate function \( g \) for \( x = 3 \) and \( x = 0 \), if possible, given that \( g \) is defined by:

\[ g(x) = \sqrt{x - 3} \]

Solution:

To find \( g(3) \), substitute \( x \) by \( 3 \) in the formula of the function:

\[ g(3) = \sqrt{3 - 3} = \sqrt{0} = 0 \]

The domain of \( g \) requires the radicand to be non-negative (\( x - 3 \geq 0 \)), giving the interval:

\[ [3, +\infty) \]

Since \( x = 0 \) is not included in the domain:

\[ g(0) = \sqrt{0 - 3} = \sqrt{-3} = \text{not a real number} \]

Evaluate, if possible, \( h(4) \), \( g(4) \), and \( \dfrac{h(4)}{g(4)} \) where functions \( h \) and \( g \) are defined by:

\[ h(x) = 3x - 8 \quad \text{and} \quad g(x) = x^2 - 16 \]

Solution:

Evaluate \( h(4) \):

\[ h(4) = 3(4) - 8 = 12 - 8 = 4 \]

Evaluate \( g(4) \):

\[ g(4) = 4^2 - 16 = 16 - 16 = 0 \]

When evaluating \( \dfrac{h(4)}{g(4)} \), the denominator \( g(4) \) is equal to \( 0 \). Division by zero is undefined in mathematics. Hence:

\[ \dfrac{h(4)}{g(4)} = \text{undefined} \]

Evaluate, if possible, \( h(t - 1) \) where function \( h \) is defined by:

\[ h(x) = 2x^2 - 2x + 2 \]

Solution:

The domain of this polynomial function is the set of all real numbers. Substitute \( t - 1 \) for \( x \):

\[ h(t - 1) = 2(t - 1)^2 - 2(t - 1) + 2 \]

Expand the square and group like terms:

\[ h(t - 1) = 2(t^2 - 2t + 1) - 2t + 2 + 2 \] \[ = 2t^2 - 4t + 2 - 2t + 4 \] \[ = 2t^2 - 6t + 6 \]

1. Solution to Exercise 1: \[ f(9) = 2(9)^2 + 2 = 2(81) + 2 = 162 + 2 = 164 \]
2. Solution to Exercise 2: \[ g(1) = 1^3 + 1 = 1 + 1 = 2 \] \[ h(1) = 1 - 1 = 0 \] \[ \dfrac{g(1)}{h(1)} = \dfrac{2}{0} = \text{undefined} \]
3. Solution to Exercise 3: \[ f(t + 2) = -2(t + 2)^2 + 2(t + 2) \] \[ = -2(t^2 + 4t + 4) + 2t + 4 \] \[ = -2t^2 - 8t - 8 + 2t + 4 \] \[ = -2t^2 - 6t - 4 \]