Answer :
Let's analyze the given functions one by one to determine their ranges and identify which one has the range [tex]\(\{y \mid y \leq 5\}\)[/tex].
1. Function [tex]\( f(x) = (x-4)^2 + 5 \)[/tex]
- This is a quadratic function that opens upwards (since the coefficient of [tex]\( (x-4)^2 \)[/tex] is positive).
- The vertex of this parabola is at [tex]\( (4, 5) \)[/tex]. Since it opens upwards, the minimum value of [tex]\( f(x) \)[/tex] is [tex]\( 5 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 4 \)[/tex], the value of [tex]\( f(x) \)[/tex] increases.
- Therefore, the range of this function is [tex]\( [5, \infty) \)[/tex].
2. Function [tex]\( f(x) = -(x-4)^2 + 5 \)[/tex]
- This is a quadratic function that opens downwards (since the coefficient of [tex]\( (x-4)^2 \)[/tex] is negative).
- The vertex of this parabola is at [tex]\( (4, 5) \)[/tex]. Since it opens downwards, the maximum value of [tex]\( f(x) \)[/tex] is [tex]\( 5 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 4 \)[/tex], the value of [tex]\( f(x) \)[/tex] decreases.
- Therefore, the range of this function is [tex]\( (-\infty, 5] \)[/tex].
3. Function [tex]\( f(x) = (x-5)^2 + 4 \)[/tex]
- This is a quadratic function that opens upwards (since the coefficient of [tex]\( (x-5)^2 \)[/tex] is positive).
- The vertex of this parabola is at [tex]\( (5, 4) \)[/tex]. Since it opens upwards, the minimum value of [tex]\( f(x) \)[/tex] is [tex]\( 4 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 5 \)[/tex], the value of [tex]\( f(x) \)[/tex] increases.
- Therefore, the range of this function is [tex]\( [4, \infty) \)[/tex].
4. Function [tex]\( f(x) = -(x-5)^2 + 4 \)[/tex]
- This is a quadratic function that opens downwards (since the coefficient of [tex]\( (x-5)^2 \)[/tex] is negative).
- The vertex of this parabola is at [tex]\( (5, 4) \)[/tex]. Since it opens downwards, the maximum value of [tex]\( f(x) \)[/tex] is [tex]\( 4 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 5 \)[/tex], the value of [tex]\( f(x) \)[/tex] decreases.
- Therefore, the range of this function is [tex]\( (-\infty, 4] \)[/tex].
Given these analyses, the function which has the range [tex]\(\{y \mid y \leq 5\}\)[/tex] is:
[tex]\[ f(x) = -(x-4)^2 + 5 \][/tex]
So, the function is:
[tex]\[ \boxed{f(x) = -(x-4)^2 + 5} \][/tex]
1. Function [tex]\( f(x) = (x-4)^2 + 5 \)[/tex]
- This is a quadratic function that opens upwards (since the coefficient of [tex]\( (x-4)^2 \)[/tex] is positive).
- The vertex of this parabola is at [tex]\( (4, 5) \)[/tex]. Since it opens upwards, the minimum value of [tex]\( f(x) \)[/tex] is [tex]\( 5 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 4 \)[/tex], the value of [tex]\( f(x) \)[/tex] increases.
- Therefore, the range of this function is [tex]\( [5, \infty) \)[/tex].
2. Function [tex]\( f(x) = -(x-4)^2 + 5 \)[/tex]
- This is a quadratic function that opens downwards (since the coefficient of [tex]\( (x-4)^2 \)[/tex] is negative).
- The vertex of this parabola is at [tex]\( (4, 5) \)[/tex]. Since it opens downwards, the maximum value of [tex]\( f(x) \)[/tex] is [tex]\( 5 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 4 \)[/tex], the value of [tex]\( f(x) \)[/tex] decreases.
- Therefore, the range of this function is [tex]\( (-\infty, 5] \)[/tex].
3. Function [tex]\( f(x) = (x-5)^2 + 4 \)[/tex]
- This is a quadratic function that opens upwards (since the coefficient of [tex]\( (x-5)^2 \)[/tex] is positive).
- The vertex of this parabola is at [tex]\( (5, 4) \)[/tex]. Since it opens upwards, the minimum value of [tex]\( f(x) \)[/tex] is [tex]\( 4 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 5 \)[/tex], the value of [tex]\( f(x) \)[/tex] increases.
- Therefore, the range of this function is [tex]\( [4, \infty) \)[/tex].
4. Function [tex]\( f(x) = -(x-5)^2 + 4 \)[/tex]
- This is a quadratic function that opens downwards (since the coefficient of [tex]\( (x-5)^2 \)[/tex] is negative).
- The vertex of this parabola is at [tex]\( (5, 4) \)[/tex]. Since it opens downwards, the maximum value of [tex]\( f(x) \)[/tex] is [tex]\( 4 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 5 \)[/tex], the value of [tex]\( f(x) \)[/tex] decreases.
- Therefore, the range of this function is [tex]\( (-\infty, 4] \)[/tex].
Given these analyses, the function which has the range [tex]\(\{y \mid y \leq 5\}\)[/tex] is:
[tex]\[ f(x) = -(x-4)^2 + 5 \][/tex]
So, the function is:
[tex]\[ \boxed{f(x) = -(x-4)^2 + 5} \][/tex]