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Let $f(x) = \min \{10 + 4x - x^2, 2x + 2\}$ for all real values of $x$. The maximum value of $f(x)$ is:

(A) 6

(B) 10

(C) 14

(D) There is no maximum value.

The correct option is (B)

Detailed Breakdown & Solution

To find the maximum value of the function $f(x) = \min \{10 + 4x - x^2, 2x + 2\}$, we need to understand how the two component functions behave and where they intersect.

Let the two functions be:

$g(x) = 10 + 4x - x^2$ (a downward-opening parabola)

$h(x) = 2x + 2$ (a straight line with a positive slope)

Step 1: Find the points of intersection

To find where $g(x)$ and $h(x)$ meet, we equate them:

$$10 + 4x - x^2 = 2x + 2$$

$$x^2 - 2x - 8 = 0$$

$$(x - 4)(x + 2) = 0$$

This gives us two intersection points: $x = 4$ and $x = -2$.

At $x = 4$, the height is $h(4) = 2(4) + 2 = 10$.

At $x = -2$, the height is $h(-2) = 2(-2) + 2 = -2$.

Step 2: Analyze the behavior of the minimum function

The function $f(x)$ takes the smaller value of the two functions at any given point $x$. Let's divide the real number line into three regions based on our intersection points:

1. Region I: $x < -2$

Let's test $x = -3$:

$g(-3) = 10 + 4(-3) - (-3)^2 = 10 - 12 - 9 = -11$

$h(-3) = 2(-3) + 2 = -4$

Since -11 is smaller than -4, $f(x)$ follows the parabola $g(x)$. Since the vertex of this parabola is at $x = 2$, this branch is strictly increasing up to $x = -2$, reaching a value of -2.

2. Region II: $-2 \le x \le 4$

Let's test $x = 2$ (the vertex of the parabola):

$g(2) = 10 + 4(2) - 2^2 = 14$

$h(2) = 2(2) + 2 = 6$

Since 6 is smaller than 14, $f(x)$ follows the straight line $h(x) = 2x + 2$. This line strictly increases from -2 (at $x = -2$) to 10 (at $x = 4$).

3. Region III: $x > 4$

Let's test $x = 5$:

$g(5) = 10 + 4(5) - 5^2 = 5$

$h(5) = 2(5) + 2 = 12$

Since 5 is smaller than 12, $f(x)$ follows the parabola $g(x)$. Since the vertex is at $x = 2$, for $x > 4$, the parabola is strictly decreasing. Its values will always be less than 10.

Conclusion

Tracing the function from left to right:

It increases on the parabolic curve up to the point $(-2, -2)$.

It then increases linearly along the line $y = 2x + 2$ from $(-2, -2)$ up to $(4, 10)$.

After $(4, 10)$, it falls along the parabolic curve toward $-\infty$.

Therefore, the absolute highest peak of the function occurs at $x = 4$, giving a maximum value of 10.