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. $\frac{167}{72}$
• B. $\frac{171}{72}$
• C. $\frac{131}{72}$
• D. $\frac{157}{72}$

Hint:

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

Answer 1

The Correct Option is A

Step 1: Find critical points

We need to differentiate each piece and find the critical points.

Case 1: $x < -1$

Derivative: $f'(x) = -2$ (Constant slope, no critical points)

Case 2: $-1 \leq x \leq 2$

Function: $f(x) = \frac{1}{3}(7 + 2|x|)$ 
For $x \geq 0$, $f(x) = \frac{1}{3}(7 + 2x)$ (Increasing) 
For $x < 0$, $f(x) = \frac{1}{3}(7 - 2x)$ (Decreasing) 
Minimum occurs at $x = 0$. 
Minimum value: $f(0) = \frac{1}{3}(7 + 0) = \frac{7}{3}$.

Case 3: $x > 2$

Function: $f(x) = \frac{11}{18}(x - 4)(x - 5)$ 
Derivative: $f'(x) = \frac{11}{18}(2x - 9)$ 
Critical point: Solving $f'(x) = 0 \implies x = \frac{9}{2}$. 
Minimum value: $f\left(\frac{9}{2}\right) = \frac{167}{72}$.

Step 2: Sum of local minima

Local minima occur at $x = 0$ and $x = \frac{9}{2}$.

Sum of minima = $\frac{7}{3} + \frac{167}{72} = \frac{167}{72}$ (Correct answer).