Determine all the critical points for the function `f(x)=6x^(5)+33x^(4)-30x^(3)+100`

Answer

To determine all the critical points for the function \( f(x) = 6x^5 + 33x^4 - 30x^3 + 100 \), we need to follow these steps:

### Step 1: Find the derivative of the function
To find the critical points, we first need to compute the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(6x^5) + \frac{d}{dx}(33x^4) - \frac{d}{dx}(30x^3) + \frac{d}{dx}(100) \]

Calculating each term, we have:

\[ f'(x) = 30x^4 + 132x^3 - 90x^2 + 0 \]

Thus,

\[ f'(x) = 30x^4 + 132x^3 - 90x^2 \]

### Step 2: Factor the derivative
Next, we can factor out the common terms in the derivative:

\[ f'(x) = 30x^2(x^2 + 4.4x - 3) \]

### Step 3: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:

\[ 30x^2(x^2 + 4.4x - 3) = 0 \]

This gives us two factors to consider:

1. \( 30x^2 = 0 \)
2. \( x^2 + 4.4x - 3 = 0 \)

### Step 4: Solve for \( x \)
From the first factor:

\[ 30x^2 = 0 \implies x = 0 \]

From the second factor, we can use the quadratic formula to find the roots:

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

Here, \( a = 1 \), \( b = 4.4 \), and \( c = -3 \):

\[ x = \frac{-4.4 \pm \sqrt{(4.4)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ (4.4)^2 - 4 \cdot 1 \cdot (-3) = 19.36 + 12 = 31.36 \]

Now substituting back into the quadratic formula:

\[ x = \frac{-4.4 \pm \sqrt{31.36}}{2} \]

Calculating \( \sqrt{31.36} \approx 5.6 \):

\[ x = \frac{-4.4 \pm 5.6}{2} \]

This gives us two solutions:

1. \( x = \frac{1.2}{2} = 0.6 \)
2. \( x = \frac{-10}{2} = -5 \)

### Step 5: List all critical points
Thus, the critical points are:

\[ x = 0, \quad x = -5, \quad x = 0.6 \]

### Final Answer
The critical points of the function \( f(x) = 6x^5 + 33x^4 - 30x^3 + 100 \) are \( x = 0, x = -5, x = 0.6 \).

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