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The problem asks to find the value of \( f(1) \), given the following conditions:

1. \( f(x) = ax^2 + bx - 5 \), where \( a \) and \( b \) are constants.
2. \( f(x-1) - f(x) = 4x - 3 \) for all real numbers \( x \).

### Step 1: Express \( f(x-1) \)

From \( f(x) = ax^2 + bx - 5 \), let's find \( f(x-1) \):

\[
f(x-1) = a(x-1)^2 + b(x-1) - 5
\]
Expand the terms:

\[
f(x-1) = a(x^2 - 2x + 1) + b(x-1) - 5
\]
\[
f(x-1) = ax^2 - 2ax + a + bx - b - 5
\]

### Step 2: Compute \( f(x-1) - f(x) \)

Now, let's subtract \( f(x) \) from \( f(x-1) \):

\[
f(x-1) - f(x) = (ax^2 - 2ax + a + bx - b - 5) - (ax^2 + bx - 5)
\]
Simplify the expression:

\[
f(x-1) - f(x) = ax^2 - 2ax + a + bx - b - 5 - ax^2 - bx + 5
\]
\[
f(x-1) - f(x) = -2ax + a - b
\]

### Step 3: Set the expression equal to \( 4x - 3 \)

We know that \( f(x-1) - f(x) = 4x - 3 \). So,

\[
-2ax + a - b = 4x - 3
\]

### Step 4: Solve for \( a \) and \( b \)

To find \( a \) and \( b \), compare the coefficients of \( x \) and the constant terms on both sides.

- For the \( x \)-terms:
  \[
  -2a = 4 \quad \Rightarrow \quad a = -2
  \]
  
- For the constant terms:
  \[
  a - b = -3 \quad \Rightarrow \quad -2 - b = -3 \quad \Rightarrow \quad b = 1
  \]

### Step 5: Find \( f(1) \)

Now that we know \( a = -2 \) and \( b = 1 \), we can substitute these values into the expression for \( f(x) \):

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

To find \( f(1) \):

\[
f(1) = -2(1)^2 + 1 - 5 = -2 + 1 - 5 = -6
\]

Thus, the value of \( f(1) \) is \( \boxed{-6} \).

Would you like a detailed explanation of any step or have further questions?

Here are five related questions for further exploration:
1. How would the result change if the equation for \( f(x-1) - f(x) \) were different?
2. Can you generalize this method for any quadratic function?
3. How would the solution change if the right-hand side of the equation were \( 4x - c \), where \( c \) is a constant?
4. What is the significance of the relationship \( f(x-1) - f(x) \)?
5. How would you approach this if the equation for \( f(x) \) was cubic instead of quadratic?

**Tip**: Always expand and simplify both sides of an equation when working with functional relationships.

Let f(x) = ax^2 + bx - 5 (a, b are constants). If f(x-1) - f(x) = 4x - 3 for all real numbers x, what is the value of f(1)?

Learn how to find the value of f(1) for a quadratic function where f(x-1) - f(x) = 4x - 3. This detailed step-by-step solution covers key algebraic concepts including function manipulation and solving quadratic functions. Suitable for students in Grades 10-12.

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