Let `f(x)={{:(,x^(3)+x^(2)-10x,-1 le x lt 0),(,cos x,0 le x lt pi/2),(,1+sin x,pi/2 le x le pi):}" Then at "x=pi/2`, f(x) has:

Options

A. a local minimum
B. a local maximum
C. no extremum
D. no local maximum

Answer

To solve the problem, we need to evaluate the function \( f(x) \) at the point \( x = \frac{\pi}{2} \) and analyze its properties. The function \( f(x) \) is defined piecewise as follows:

\[ f(x) = \begin{cases} x^3 + x^2 - 10x & \text{for } -1 \leq x < 0 \\ \cos x & \text{for } 0 \leq x < \frac{\pi}{2} \\ 1 + \sin x & \text{for } \frac{\pi}{2} \leq x \leq \pi \end{cases} \]

### Step 1: Identify the relevant piece of the function at \( x = \frac{\pi}{2} \)

Since \( x = \frac{\pi}{2} \) falls in the interval \( \frac{\pi}{2} \leq x \leq \pi \), we will use the third piece of the function:

\[ f(x) = 1 + \sin x \]

### Step 2: Evaluate \( f\left(\frac{\pi}{2}\right) \)

Now we need to compute \( f\left(\frac{\pi}{2}\right) \):

\[ f\left(\frac{\pi}{2}\right) = 1 + \sin\left(\frac{\pi}{2}\right) \]

Since \( \sin\left(\frac{\pi}{2}\right) = 1 \):

\[ f\left(\frac{\pi}{2}\right) = 1 + 1 = 2 \]

### Step 3: Analyze the continuity of \( f(x) \) at \( x = \frac{\pi}{2} \)

To determine if \( f(x) \) has any special properties at \( x = \frac{\pi}{2} \), we need to check the left-hand limit and right-hand limit at this point.

**Left-hand limit as \( x \to \frac{\pi}{2}^- \)**:

Using the second piece of the function \( f(x) = \cos x \):

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \cos\left(\frac{\pi}{2}\right) = 0 \]

**Right-hand limit as \( x \to \frac{\pi}{2}^+ \)**:

Using the third piece of the function \( f(x) = 1 + \sin x \):

\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = 1 + \sin\left(\frac{\pi}{2}\right) = 2 \]

### Step 4: Conclusion about continuity and differentiability

Since the left-hand limit (0) does not equal the right-hand limit (2), the function \( f(x) \) is not continuous at \( x = \frac{\pi}{2} \). Therefore, \( f(x) \) does not have a derivative at this point, and we can conclude that:

At \( x = \frac{\pi}{2} \), \( f(x) \) has a jump discontinuity.

### Final Answer

At \( x = \frac{\pi}{2} \), \( f(x) \) has a jump discontinuity.

---