Find the points of local maxima and local minima, if any, of the following functions. Find also the local maximum and local minimum values :
`f(x)=-(3)/(4)x^(4)-8x^(3)-(45)/(2)x^(2)+105`

Answer

To find the points of local maxima and minima for the function \( f(x) = -\frac{3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 + 105 \), we will follow these steps:

### Step 1: Find the first derivative of the function
We need to find the first derivative \( f'(x) \) and set it to zero to find critical points.

\[ f'(x) = -3x^3 - 24x^2 - 45x \]

### Step 2: Set the first derivative to zero
Now we set the first derivative equal to zero to find the critical points:

\[ -3x^3 - 24x^2 - 45x = 0 \]

Factoring out \(-3x\):

\[ -3x(x^2 + 8x + 15) = 0 \]

This gives us:

\[ x = 0 \quad \text{or} \quad x^2 + 8x + 15 = 0 \]

### Step 3: Solve the quadratic equation
Now we solve the quadratic equation \( x^2 + 8x + 15 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = 8, c = 15 \):

\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 15}}{2 \cdot 1} \]
\[ x = \frac{-8 \pm \sqrt{64 - 60}}{2} \]
\[ x = \frac{-8 \pm \sqrt{4}}{2} \]
\[ x = \frac{-8 \pm 2}{2} \]

This gives us:

\[ x = -3 \quad \text{and} \quad x = -5 \]

### Step 4: Identify critical points
The critical points are:

\[ x = 0, \quad x = -3, \quad x = -5 \]

### Step 5: Find the second derivative
Next, we find the second derivative \( f''(x) \):

\[ f''(x) = -9x^2 - 48x - 45 \]

### Step 6: Evaluate the second derivative at critical points
Now we evaluate \( f''(x) \) at each critical point to determine if they are maxima or minima.

1. **At \( x = 0 \)**:
\[ f''(0) = -9(0)^2 - 48(0) - 45 = -45 \quad (\text{negative, local maximum}) \]

2. **At \( x = -3 \)**:
\[ f''(-3) = -9(-3)^2 - 48(-3) - 45 = -81 + 144 - 45 = 18 \quad (\text{positive, local minimum}) \]

3. **At \( x = -5 \)**:
\[ f''(-5) = -9(-5)^2 - 48(-5) - 45 = -225 + 240 - 45 = -30 \quad (\text{negative, local maximum}) \]

### Step 7: Calculate the function values at critical points
Now we calculate the function values at the critical points to find the local maxima and minima.

1. **At \( x = 0 \)**:
\[ f(0) = 105 \]

2. **At \( x = -3 \)**:
\[ f(-3) = -\frac{3}{4}(-3)^4 - 8(-3)^3 - \frac{45}{2}(-3)^2 + 105 \]
\[ = -\frac{3}{4}(81) + 216 - \frac{45}{2}(9) + 105 \]
\[ = -60.75 + 216 - 202.5 + 105 = 57.75 \]

3. **At \( x = -5 \)**:
\[ f(-5) = -\frac{3}{4}(-5)^4 - 8(-5)^3 - \frac{45}{2}(-5)^2 + 105 \]
\[ = -\frac{3}{4}(625) + 1000 - \frac{45}{2}(25) + 105 \]
\[ = -468.75 + 1000 - 562.5 + 105 = 73.75 \]

### Summary of Results
- Local maximum at \( x = 0 \) with value \( f(0) = 105 \).
- Local minimum at \( x = -3 \) with value \( f(-3) = 57.75 \).
- Local maximum at \( x = -5 \) with value \( f(-5) = 73.75 \).