Key Concepts and Formulas

• Local Minimum: A point $x=c$ is a local minimum of a function $f(x)$ if $f(c) \le f(x)$ for all $x$ in some open interval containing $c$.
• First Derivative Test: If $f'(x)$ changes from negative to positive at $x=c$, then $f(x)$ has a local minimum at $x=c$.
• Piecewise Functions: Analyze each piece separately and also check the points where the function definition changes.

Step-by-Step Solution

Step 1: Analyze the first piece of the function, $f(x) = 1 - 2x$ for $x < -1$.

Since $f(x) = 1 - 2x$ is a linear function with a negative slope, it is strictly decreasing for $x < -1$. Therefore, there is no local minimum in this interval.

Step 2: Analyze the second piece of the function, $f(x) = \frac{1}{3}(7 + 2|x|)$ for $-1 \leq x \leq 2$.

We can rewrite this piece as: $f(x) = \begin{cases} \frac{1}{3}(7 - 2x), & -1 \leq x < 0 \\ \frac{1}{3}(7 + 2x), & 0 \leq x \leq 2 \end{cases}$

Step 3: Find the derivative of the second piece.

For $-1 \leq x < 0$, $f'(x) = -\frac{2}{3}$. For $0 < x \leq 2$, $f'(x) = \frac{2}{3}$.

At $x = 0$, the derivative changes from negative to positive. Therefore, there is a local minimum at $x = 0$. The value of the function at $x = 0$ is $f(0) = \frac{1}{3}(7 + 2|0|) = \frac{7}{3}$.

Step 4: Analyze the third piece of the function, $f(x) = \frac{11}{18}(x - 4)(x - 5)$ for $x > 2$.

Expand the expression: $f(x) = \frac{11}{18}(x^2 - 9x + 20)$

Step 5: Find the derivative of the third piece.

$f'(x) = \frac{11}{18}(2x - 9)$

Step 6: Find the critical points of the third piece by setting the derivative equal to zero.

$f'(x) = 0 \Rightarrow \frac{11}{18}(2x - 9) = 0 \Rightarrow 2x - 9 = 0 \Rightarrow x = \frac{9}{2} = 4.5$ Since $4.5 > 2$, this critical point is within the interval.

Step 7: Determine if the critical point is a local minimum using the second derivative test.

$f''(x) = \frac{11}{18}(2) = \frac{11}{9}$ Since $f''(4.5) = \frac{11}{9} > 0$, the critical point $x = 4.5$ is a local minimum.

Step 8: Calculate the value of the function at the local minimum $x = 4.5$.

$f(4.5) = \frac{11}{18}(4.5 - 4)(4.5 - 5) = \frac{11}{18}(0.5)(-0.5) = \frac{11}{18}\left(-\frac{1}{4}\right) = -\frac{11}{72}$

Step 9: Check the function values at the junction points $x = -1$ and $x = 2$.

At $x = -1$, $f(-1) = \frac{1}{3}(7 + 2|-1|) = \frac{1}{3}(7 + 2) = \frac{9}{3} = 3$. For $x < -1$, $f(x) = 1 - 2x$. As $x$ approaches $-1$ from the left, $f(x)$ approaches $1 - 2(-1) = 3$. Thus, the function is continuous at $x=-1$. Since $f'(x) = -2$ for $x < -1$ and $f'(x) = -2/3$ for $-1<x<0$, there is no local minimum at $x=-1$.

At $x = 2$, $f(2) = \frac{1}{3}(7 + 2|2|) = \frac{1}{3}(7 + 4) = \frac{11}{3}$. For $x > 2$, $f(x) = \frac{11}{18}(x - 4)(x - 5)$. As $x$ approaches $2$ from the right, $f(x)$ approaches $\frac{11}{18}(2 - 4)(2 - 5) = \frac{11}{18}(-2)(-3) = \frac{11}{18}(6) = \frac{11}{3}$. Thus, the function is continuous at $x=2$. However, we must check whether $x=2$ is a local minimum. For $x<2$, $f'(x) = 2/3 > 0$. For $x>2$, $f'(x) = \frac{11}{18}(2x-9)$. At $x=2$, $f'(2) = \frac{11}{18}(4-9) = -\frac{55}{18} < 0$. Since the derivative changes from positive to negative, $x=2$ is a local maximum, not a local minimum.

Step 10: Find the sum of all local minimum values.

The local minimum values are $\frac{7}{3}$ and $-\frac{11}{72}$. The sum is $\frac{7}{3} - \frac{11}{72} = \frac{7 \cdot 24}{3 \cdot 24} - \frac{11}{72} = \frac{168}{72} - \frac{11}{72} = \frac{157}{72}$

There is an error in the provided correct answer. The correct sum of local minimums should be $\frac{157}{72}$.

Step 11: Re-evaluating the question since the answer is not matching The function is: $\mathrm{f}(x)=\left\{\begin{array}{lr} 1-2 x, & x<-1 \\ \frac{1}{3}(7+2|x|), & -1 \leq x \leq 2 \\ \frac{11}{18}(x-4)(x-5), & x>2 \end{array}\right.$

We found local minimum at $x=0$ where $f(0) = 7/3$. Also, local minimum at $x=4.5$, where $f(4.5) = -11/72$. The sum is $\frac{7}{3} - \frac{11}{72} = \frac{168-11}{72} = \frac{157}{72}$

Let us evaluate the end points. at $x=-1$, $f(-1) = \frac{1}{3}(7+2) = \frac{9}{3} = 3$. at $x=2$, $f(2) = \frac{1}{3}(7+4) = \frac{11}{3}$.

Now let us consider $x<-1$. $f(x) = 1-2x$. As $x \to -1^-$, $f(x) \to 1-2(-1) = 3$. Since $f'(x) = -2 < 0$, this is always decreasing. No local minimum.

For $-1 \le x \le 2$, $f(x) = \frac{1}{3}(7+2|x|)$. $f'(x) = \frac{2}{3} \frac{x}{|x|}$. At $x=0$, $f(0) = \frac{7}{3}$.

For $x>2$, $f(x) = \frac{11}{18}(x-4)(x-5) = \frac{11}{18}(x^2 - 9x + 20)$. $f'(x) = \frac{11}{18}(2x-9)$. $f'(x) = 0$ at $x = 4.5$. $f(4.5) = \frac{11}{18}(0.5)(-0.5) = -\frac{11}{72}$.

The sum is $\frac{7}{3} - \frac{11}{72} = \frac{168-11}{72} = \frac{157}{72}$

Common Mistakes & Tips

• Remember to consider the points where the function definition changes, as these can be local minima or maxima.
• Be careful when dealing with absolute values; rewrite the function as a piecewise function to avoid errors.
• Always check if the critical points you find are within the specified intervals.

Summary

We analyzed the given piecewise function by considering each piece separately and the points where the definition changes. We found local minimum values at $x = 0$ and $x = 4.5$, with corresponding values of $\frac{7}{3}$ and $-\frac{11}{72}$. The sum of these local minimum values is $\frac{157}{72}$.

Final Answer The final answer is \boxed{\frac{157}{72}}, which corresponds to option (B).