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The image contains several problems where functions need to be evaluated at given points. Let’s solve each of them step-by-step:

### Problem 1: \( f(x) = 2x^3 - 3x^2 + x - 10 \)

1. **a.** Evaluate \( f(2) \):

   \[
   f(2) = 2(2)^3 - 3(2)^2 + 2 - 10 = 2 \times 8 - 3 \times 4 + 2 - 10 = 16 - 12 + 2 - 10 = -4.
   \]

2. **b.** Evaluate \( f\left(\frac{1}{2}\right) \):

   \[
   f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)^2 + \frac{1}{2} - 10 = 2 \times \frac{1}{8} - 3 \times \frac{1}{4} + \frac{1}{2} - 10 = \frac{1}{4} - \frac{3}{4} + \frac{1}{2} - 10 = -\frac{9}{4}.
   \]

3. **c.** Evaluate \( f(x + 1) \):

   \[
   f(x + 1) = 2(x + 1)^3 - 3(x + 1)^2 + (x + 1) - 10.
   \]

   Expand and simplify this expression further.

### Problem 2: \( f(x) = \frac{x^2 - x}{x + 1} \)

1. **a.** Evaluate \( f(2) \):

   \[
   f(2) = \frac{2^2 - 2}{2 + 1} = \frac{4 - 2}{3} = \frac{2}{3}.
   \]

2. **b.** Evaluate \( f(2) \): 

   Same as above: \( f(2) = \frac{2}{3}. \)

3. **c.** Evaluate \( f\left(\frac{1}{2}\right) \):

   \[
   f\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^2 - \frac{1}{2}}{\frac{1}{2} + 1} = \frac{\frac{1}{4} - \frac{1}{2}}{\frac{3}{2}} = \frac{-\frac{1}{4}}{\frac{3}{2}} = -\frac{1}{6}.
   \]

### Problem 3: \( f(x)

Evaluate the given functions:
1) f(x) = 2x^3 - 3x^2 + x - 10
  a) f(2)
  b) f(1/2)
  c) f(x + 1)

2) f(x) = (x^2 - x) / (x + 1)
  a) f(2)
  b) f(2)
  c) f(1/2)

3) f(x) = x + 2
  a) f(2)
  b) f(1/2)
  c) f(-2)

4) f(x) = x^4
  a) f(x + 3)

Step-by-step solutions for evaluating various polynomial and rational functions at specific points. Learn how to substitute and simplify complex expressions. Suitable for high school students in Grades 9-12.

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