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.