Show that f(x)=x^2-2x-3, 0**≤x≤**5 is a one-to-one function with algebraically 

To show that the function \( f(x) = x^2 - 2x - 3 \) is one-to-one over the interval \( 0 \leq x \leq 5 \), we will demonstrate that the function is strictly either increasing or decreasing in this interval.

### Step 1: Compute the derivative
To analyze whether the function is increasing or decreasing, we first compute the derivative of \( f(x) \). The derivative tells us the rate of change of the function.

\[
f'(x) = \frac{d}{dx}(x^2 - 2x - 3)
\]

Using standard rules of differentiation:

\[
f'(x) = 2x - 2
\]

### Step 2: Analyze the sign of the derivative
Now, we analyze the sign of \( f'(x) = 2x - 2 \) over the interval \( 0 \leq x \leq 5 \).

- When \( x = 0 \): 
  \[
  f'(0) = 2(0) - 2 = -2
  \]
  So, \( f'(x) < 0 \) at \( x = 0 \).
  
- When \( x = 5 \):
  \[
  f'(5) = 2(5) - 2 = 8
  \]
  So, \( f'(x) > 0 \) at \( x = 5 \).

### Step 3: Determine intervals of increase or decrease
The critical point occurs when \( f'(x) = 0 \). Setting the derivative equal to zero and solving for \( x \):

\[
2x - 2 = 0
\]
\[
x = 1
\]

Thus, \( f(x) \) has a critical point at \( x = 1 \). To determine the behavior of the function in the interval \( 0 \leq x \leq 5 \), we examine the sign of the derivative on the subintervals \( 0 \leq x < 1 \) and \( 1 < x \leq 5 \):

- For \( 0 \leq x < 1 \), \( f'(x) = 2x - 2 \) is negative (since \( x < 1 \)), meaning that the function is decreasing in this interval.
- For \( 1 < x \leq 5 \), \( f'(x) = 2x - 2 \) is positive (since \( x > 1 \)), meaning that the function is increasing in this interval.

Thus, \( f(x) \) is decreasing from \( x = 0 \) to \( x = 1 \) and increasing from \( x = 1 \) to \( x = 5 \).

### Step 4: Conclusion
Since the function is strictly decreasing on \( [0, 1] \) and strictly increasing on \( [1, 5] \), it does not have any repeating \( y \)-values in this interval. Therefore, the function \( f(x) = x^2 - 2x - 3 \) is one-to-one on \( [0, 5] \).

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### Related questions:
1. How do you use the first derivative test to classify critical points?
2. What is the difference between a one-to-one function and an onto function?
3. How can you use the second derivative to determine concavity?
4. How do we find the inverse of a one-to-one function?
5. How do you prove that a quadratic function is not one-to-one in general?

**Tip:** A function is one-to-one if and only if its derivative does not change sign, meaning it is either strictly increasing or decreasing throughout its domain.

Show that f(x)=x^2-2x-3, 0 ≤ x ≤ 5 is a one-to-one function algebraically.

Discover how to prove that the quadratic function f(x)=x^2-2x-3 is one-to-one on the interval [0,5] using algebraic methods. This step-by-step guide explains how the derivative shows where the function increases and decreases, ensuring no repeated y-values. Suitable for Grades 10-12 students.

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