### Instruction: For each value of $x,$ $f(x)$ is defined to be the minimum value of the three numbers $2x + 2,$ $\frac{1}{2} x + 1,$ and $-\frac{3}{4} x + 7.$  Find the maximum value of $f(x).$.### Response: I want to find the maximum value of $f(x),$ which means I want to find the value of $x$ that makes $f(x)$ as large as possible.
I notice that $f(x)$ is the minimum of three linear expressions, so it will change its slope whenever one of those expressions becomes equal to another.
That means I should look for the points of intersection of the three lines $y = 2x + 2,$ $y = \frac{1}{2} x + 1,$ and $y = -\frac{3}{4} x + 7.$
To find those points, I can solve for $x$ by setting each pair of equations equal to each other and simplifying.
For example, to find the point where $y = 2x + 2$ and $y = \frac{1}{2} x + 1$ intersect, I can write $2x + 2 = \frac{1}{2} x + 1$ and subtract $\frac{1}{2} x + 1$ from both sides to get $1.5x + 1 = 0.$
Then I can subtract 1 from both sides and divide by 1.5 to get $x = -\frac{2}{3}.$
This means that when $x = -\frac{2}{3},$ $f(x) = \frac{1}{2} x + 1 = -\frac{1}{3} + 1 = \frac{2}{3}.$
Similarly, I can find the point where $y = 2x + 2$ and $y = -\frac{3}{4} x + 7$ intersect by writing $2x + 2 = -\frac{3}{4} x + 7$ and solving for $x.$
I get $x = \frac{20}{11}$ and $f(x) = 2x + 2 = \frac{40}{11} + 2 = \frac{62}{11}.$
Finally, I can find the point where $y = \frac{1}{2} x + 1$ and $y = -\frac{3}{4} x + 7$ intersect by writing $\frac{1}{2} x + 1 = -\frac{3}{4} x + 7$ and solving for $x.$
I get $x = \frac{24}{5}$ and $f(x) = \frac{1}{2} x + 1 = \frac{12}{5} + 1 = \frac{17}{5}.$
Now I have three candidates for the maximum value of $f(x),$ namely $\frac{2}{3},$ $\frac{62}{11},$ and $\frac{17}{5}.$
To compare them, I can convert them to decimals or find a common denominator.
I decide to use decimals, since they are easier to compare.
I get $\frac{2}{3} \approx 0.67,$ $\frac{62}{11} \approx 5.64,$ and $\frac{17}{5} \approx 3.40.$
Clearly, the largest of these values is $\frac{62}{11},$ so that is the maximum value of $f(x).$