Question

The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \]

• A. \(\frac{157}{72}\)
• B. \(\frac{171}{72}\)
• C. \(\frac{131}{72}\)
• D. \(\frac{167}{72}\)

Hint:

Piecewise functions may have multiple local minima or maxima, examine all segments thoroughly.

Answer 1

The Correct Option is A

We need to find the local minimum values of the piecewise function \( f(x) \).

Case 1: \( x < -1 \)
\( f(x) = 1 - 2x \)
This is a decreasing linear function with no local minima.

Case 2: \( -1 \leq x \leq 2 \)
\( f(x) = \frac{1}{3}(7 + 2|x|) \)
For \( -1 \leq x \leq 0 \): \( f(x) = \frac{1}{3}(7 - 2x) \), \( f'(x) = -\frac{2}{3} \) (decreasing)
For \( 0 \leq x \leq 2 \): \( f(x) = \frac{1}{3}(7 + 2x) \), \( f'(x) = \frac{2}{3} \) (increasing)
Local minimum at \( x = 0 \): \( f(0) = \frac{7}{3} \)

Case 3: \( x > 2 \)
\( f(x) = \frac{11}{18}(x-4)(x-5) \)
Critical point at \( x = 4.5 \):
\( f(4.5) = -\frac{11}{72} \) (local minimum since \( f''(x) > 0 \))

Continuity Check at \( x = 2 \):
Both pieces give \( f(2) = \frac{11}{3} \), confirming continuity but no additional extrema.

Sum of Local Minima:
\[ \frac{7}{3} + \left(-\frac{11}{72}\right) = \frac{168}{72} - \frac{11}{72} = \frac{157}{72} \]

Final Answer:
The sum of all local minimum values is \(\dfrac{157}{72}\).