Questions: Find all the critical numbers of f(x) = (3/2) x^4 - 4 x^3 + 3 x^2 + 2, then determine the local minimum and maximum points by using a graph.

Transcript text: Find all the critical numbers of $f(x)=\frac{3}{2} x^{4}-4 x^{3}+3 x^{2}+2$, then determine the local minimum and maximum points by using a graph.

Solution

Solution Steps

To find the critical numbers of the function $ f(x) = \frac{3}{2} x^4 - 4x^3 + 3x^2 + 2 $, we need to:

Compute the first derivative $ f'(x) $.
Set $ f'(x) = 0 $ and solve for $ x $ to find the critical points.
Use the second derivative test or analyze the graph to determine the nature of these critical points (local minima, maxima, or neither).

Step 1: Find the First Derivative

To find the critical numbers of the function $ f(x) = \frac{3}{2} x^4 - 4x^3 + 3x^2 + 2 $, we first compute the first derivative: \[ f'(x) = 6x^3 - 12x^2 + 6x \]

Step 2: Solve for Critical Points

Next, we set the first derivative equal to zero and solve for $ x $: \[ 6x^3 - 12x^2 + 6x = 0 \] \[ 6x(x^2 - 2x + 1) = 0 \] \[ 6x(x - 1)^2 = 0 \] Thus, the critical points are: \[ x = 0 \quad \text{and} \quad x = 1 \]

Step 3: Determine the Nature of Critical Points

To determine the nature of these critical points, we compute the second derivative: \[ f''(x) = 18x^2 - 24x + 6 \]

We then evaluate the second derivative at each critical point:

At $ x = 0 $: \[ f''(0) = 18(0)^2 - 24(0) + 6 = 6 \] Since $ f''(0) > 0 $, $ x = 0 $ is a local minimum.
At $ x = 1 $: \[ f''(1) = 18(1)^2 - 24(1) + 6 = 0 \] Since $ f''(1) = 0 $, the second derivative test is inconclusive at $ x = 1 $.

Final Answer

The critical numbers of the function are $ x = 0 $ and $ x = 1 $. The point $ (0, 2) $ is a local minimum. The second derivative test is inconclusive at $ x = 1 $.

\[ \boxed{\text{Critical numbers: } x = 0 \text{ and } x = 1} \] \[ \boxed{(0, 2) \text{ is a local minimum}} \]

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