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Let's solve the problems presented in the uploaded image one by one.

### 1. Evaluate the following functions at \( x = -1 \):
(a) \( f(x) = x^2 - 4x - 64 \)

\[
f(-1) = (-1)^2 - 4(-1) - 64 = 1 + 4 - 64 = -59
\]

(b) \( g(x) = 3x^3 - x^2 + 3x - 1 \)

\[
g(-1) = 3(-1)^3 - (-1)^2 + 3(-1) - 1 = -3 - 1 - 3 - 1 = -8
\]

(c) \( h(x) = \frac{\sqrt{5 - x} - 3}{x + 1} \)

\[
h(-1) = \frac{\sqrt{5 - (-1)} - 3}{-1 + 1} = \frac{\sqrt{6} - 3}{0}
\]

Since the denominator is zero, \( h(-1) \) is undefined.

### 2. Given \( f(x) = \begin{cases} 
\frac{x - 2}{x} & \text{if } x < 2 \\
\frac{x}{x - 2} & \text{if } x \geq 2 
\end{cases} \), give the values of the following:
(a) \( f(2) \)

\[
f(2) = \frac{2}{2 - 2} = \frac{2}{0}
\]

Since the denominator is zero, \( f(2) \) is undefined.

(b) \( f(12.5) \)

\[
f(12.5) = \frac{12.5}{12.5 - 2} = \frac{12.5}{10.5} \approx 1.19
\]

(c) \( f(-0.5) \)

\[
f(-0.5) = \frac{-0.5 - 2}{-0.5} = \frac{-2.5}{-0.5} = 5
\]

(d) \( f(5) \)

\[
f(5) = \frac{5}{5 - 2} = \frac{5}{3} \approx 1.67
\]

(e) \( f(0.5) \)

\[
f(0.5) = \frac{0.5 - 2}{0.5} = \frac{-1.5}{0.5} = -3
\]

### 3. Given \( f(x) = x^3 - 4x + 1 \), solve for \( f(1.5) \)
(a) Solve for \( f(1.5) \)

\[
f(1.5) = (1.5)^3 - 4(1.5) + 1 = 3.375 - 6 + 1 = -1.625
\]

(b) Is \( f(x + 1) \) the same as \( f(x) + f(1) \)?

\[
f(x + 1) = (x + 1)^3 - 4(x + 1) + 1 = x^3 + 3x^2 + 3x + 1 - 4x - 4 + 1 = x^3 + 3x^2 - x - 2
\]

\[
f(x) + f(1) = (x^3 - 4x + 1) + (1 - 4 + 1) = x^3 - 4x + 1 - 2 = x^3 - 4x - 1
\]

Since \( f(x + 1) \neq f(x) + f(1) \), they are not the same.

Would you like more details or have any questions about the solutions? Here are some related questions for further practice:

1. Evaluate \( f(x) = x^3 - 2x^2 + 5x - 7 \) at \( x = -2 \).
2. Solve \( g(x) = \frac{3x + 2}{x - 1} \) for \( x = 0 \).
3. Determine the limit of \( h(x) = \frac{\sqrt{x + 4} - 2}{x} \) as \( x \) approaches 0.
4. Find the derivative of \( f(x) = x^2 - 4x + 4 \) and evaluate it at \( x = 3 \).
5. Solve the integral \( \int (3x^2 - 4x + 1) \, dx \).
6. Given \( f(x) = e^x \), find the value of \( f''(0) \).
7. Determine the roots of the equation \( x^2 - 6x + 9 = 0 \).
8. Find the inverse of the function \( f(x) = 2x + 3 \).

**Tip:** Always check the domain of the function before evaluating it at a specific point to avoid undefined values.

Explore solutions to mathematical functions including evaluations and definitions. Solve problems involving functions such as quadratic and piecewise functions. Suitable for Grades 10-12.

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