To find the range of the function [tex]\( f(x) = x^2 - 2x + 5 \)[/tex] on the interval [tex]\( (0, 3] \)[/tex], we need to analyze its behavior on this interval step-by-step. Here’s the detailed solution:
1. Evaluate the function at the endpoints of the interval [tex]\( (0, 3] \)[/tex]:
- At [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = 0^2 - 2 \cdot 0 + 5 = 5
\][/tex]
- At [tex]\( x = 3 \)[/tex]:
[tex]\[
f(3) = 3^2 - 2 \cdot 3 + 5 = 9 - 6 + 5 = 8
\][/tex]
2. Determine the critical points within the interval [tex]\( (0, 3] \)[/tex]:
- Find the derivative of the function to locate the critical points:
[tex]\[
f'(x) = \frac{d}{dx}(x^2 - 2x + 5) = 2x - 2
\][/tex]
- Set the derivative equal to zero to find the critical points:
[tex]\[
2x - 2 = 0 \implies x = 1
\][/tex]
- Check if the critical point [tex]\( x = 1 \)[/tex] lies in the interval [tex]\( (0, 3] \)[/tex]:
Since [tex]\( 1 \)[/tex] is in the interval [tex]\( (0, 3] \)[/tex], it is a valid critical point.
3. Evaluate the function at the critical point:
- At [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = 1^2 - 2 \cdot 1 + 5 = 1 - 2 + 5 = 4
\][/tex]
4. Identify the range of the function:
- We now have the function values at the lower endpoint [tex]\( (0) \)[/tex], upper endpoint [tex]\( (3) \)[/tex], and the critical point [tex]\( (1) \)[/tex]:
[tex]\[
f(0) = 5, \quad f(1) = 4, \quad f(3) = 8
\][/tex]
- By considering these values, the smallest function value is [tex]\( 4 \)[/tex] and the largest is [tex]\( 8 \)[/tex].
Thus, the range [tex]\( A \)[/tex] of the function [tex]\( f(x) = x^2 - 2x + 5 \)[/tex] on the interval [tex]\( (0, 3] \)[/tex] is:
[tex]\[
A = [4, 8]
\][/tex]
