Show that there lies a point on the curve `f(x)=x(x+3)e^(-pi/2),-3lexle0` where tangent drawn is parallel to the x-axis.

Answer

To solve the problem, we need to show that there exists a point on the curve \( f(x) = x(x + 3)e^{-\frac{\pi}{2}} \) within the interval \([-3, 0]\) where the tangent drawn is parallel to the x-axis. This means we need to find a point where the derivative \( f'(x) \) is equal to 0.

### Step 1: Find the derivative of \( f(x) \)

The function is given as:
\[ f(x) = x(x + 3)e^{-\frac{\pi}{2}} \]

To find the derivative \( f'(x) \), we will use the product rule. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then:
\[ (uv)' = u'v + uv' \]

Let:
- \( u = x \)
- \( v = (x + 3)e^{-\frac{\pi}{2}} \)

Now, we need to find \( u' \) and \( v' \):
- \( u' = 1 \)
- \( v = (x + 3)e^{-\frac{\pi}{2}} \) (here \( e^{-\frac{\pi}{2}} \) is a constant)

Using the product rule:
\[ f'(x) = u'v + uv' = 1 \cdot (x + 3)e^{-\frac{\pi}{2}} + x \cdot (1)e^{-\frac{\pi}{2}} \]
\[ = (x + 3)e^{-\frac{\pi}{2}} + xe^{-\frac{\pi}{2}} = (2x + 3)e^{-\frac{\pi}{2}} \]

### Step 2: Set the derivative equal to zero

To find where the tangent is parallel to the x-axis, we set the derivative equal to zero:
\[ f'(x) = (2x + 3)e^{-\frac{\pi}{2}} = 0 \]

Since \( e^{-\frac{\pi}{2}} \) is a constant and never zero, we can ignore it:
\[ 2x + 3 = 0 \]

### Step 3: Solve for \( x \)

Now, we solve for \( x \):
\[ 2x = -3 \implies x = -\frac{3}{2} \]

### Step 4: Check if \( x \) is within the interval

Now we need to check if \( x = -\frac{3}{2} \) lies within the interval \([-3, 0]\):
\[ -3 \leq -\frac{3}{2} \leq 0 \]

This is true, so \( x = -\frac{3}{2} \) is indeed within the interval.

### Conclusion

Thus, we have shown that there exists a point on the curve \( f(x) \) where the tangent is parallel to the x-axis at \( x = -\frac{3}{2} \).

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