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

Local minimum values of the piecewise function \( f(x) \) are determined as follows.

Case 1: \( x < -1 \)
\( f(x) = 1 - 2x \)
This linear function is strictly decreasing and thus possesses no local minima.

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

Case 3: \( x > 2 \)
\( f(x) = \frac{11}{18}(x-4)(x-5) \)
The critical point is at \( x = 4.5 \).
\( f(4.5) = -\frac{11}{72} \) is a local minimum because \( f''(x)>0 \).

Continuity Check at \( x = 2 \):
At \( x = 2 \), \( f(2) = \frac{11}{3} \) for both function definitions, confirming continuity. No additional extrema are introduced at this point.

Sum of Local Minima:
The sum of the identified local minima is calculated as:
\[ \frac{7}{3} + \left(-\frac{11}{72}\right) = \frac{168}{72} - \frac{11}{72} = \frac{157}{72} \]

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