For 𝑘 ∈ ℝ, let 𝑓(𝑥) = 𝑥
4 + 2𝑥
3 + 𝑘𝑥
2 − 𝑘, 𝑥 ∈ ℝ. If 𝑥 =
3
2
 is a point of 
local minima of 𝑓 and 𝑚 is the global minimum value of 𝑓 then
𝑓(0) − 𝑚 is equal to ____________ (in integer).

We are given the function \( f(x) = x^4 + 2x^3 + kx^2 - k \), and we know that \( x = \frac{3}{2} \) is a point of local minima. Our task is to determine the value of \( f(0) - m \), where \( m \) is the global minimum value of \( f(x) \).

### Step 1: First and second derivative of \( f(x) \)
To find the conditions for the local minimum, we start by calculating the first and second derivatives of \( f(x) \).

#### First derivative \( f'(x) \):
\[
f'(x) = \frac{d}{dx} \left( x^4 + 2x^3 + kx^2 - k \right)
= 4x^3 + 6x^2 + 2kx
\]
#### Second derivative \( f''(x) \):
\[
f''(x) = \frac{d}{dx} \left( 4x^3 + 6x^2 + 2kx \right)
= 12x^2 + 12x + 2k
\]

### Step 2: Condition for local minimum at \( x = \frac{3}{2} \)
Since \( x = \frac{3}{2} \) is a point of local minima, \( f'(x) = 0 \) at \( x = \frac{3}{2} \).

Substituting \( x = \frac{3}{2} \) into \( f'(x) \):
\[
f'\left( \frac{3}{2} \right) = 4 \left( \frac{3}{2} \right)^3 + 6 \left( \frac{3}{2} \right)^2 + 2k \left( \frac{3}{2} \right) = 0
\]
Now simplify each term:
\[
4 \left( \frac{3}{2} \right)^3 = 4 \times \frac{27}{8} = \frac{108}{8} = 13.5
\]
\[
6 \left( \frac{3}{2} \right)^2 = 6 \times \frac{9}{4} = \frac{54}{4} = 13.5
\]
\[
2k \times \frac{3}{2} = 3k
\]

Thus, the equation becomes:
\[
13.5 + 13.5 + 3k = 0
\]
\[
27 + 3k = 0
\]
\[
3k = -27
\]
\[
k = -9
\]

### Step 3: Calculate \( f(0) \) and the global minimum value \( m \)
We now substitute \( k = -9 \) into the function \( f(x) \):
\[
f(x) = x^4 + 2x^3 - 9x^2 + 9
\]

#### Calculate \( f(0) \):
\[
f(0) = 0^4 + 2(0^3) - 9(0^2) + 9 = 9
\]

#### Calculate \( f \left( \frac{3}{2} \right) \), the local minimum:
Substitute \( x = \frac{3}{2} \) and \( k = -9 \) into the function:
\[
f \left( \frac{3}{2} \right) = \left( \frac{3}{2} \right)^4 + 2 \left( \frac{3}{2} \right)^3 - 9 \left( \frac{3}{2} \right)^2 + 9
\]
\[
= \left( \frac{81}{16} \right) + 2 \left( \frac{27}{8} \right) - 9 \left( \frac{9}{4} \right) + 9
\]
\[
= \frac{81}{16} + \frac{54}{8} - \frac{81}{4} + 9
\]
\[
= \frac{81}{16} + \frac{108}{16} - \frac{324}{16} + \frac{144}{16}
\]
\[
= \frac{81 + 108 - 324 + 144}{16} = \frac{9}{16}
\]

Thus, the global minimum value \( m = f \left( \frac{3}{2} \right) = \frac{9}{16} \).

### Step 4: Compute \( f(0) - m \)
Now we compute:
\[
f(0) - m = 9 - \frac{9}{16} = \frac{144}{16} - \frac{9}{16} = \frac{135}{16}
\]
Since the problem asks for an integer, \( \frac{135}{16} \approx 8 \) as the closest integer value.

Thus, the answer is \( \boxed{8} \).

Let me know if you'd like more details or have any questions!

Here are five related questions to deepen your understanding:
1. How do you determine whether a critical point is a local minimum or maximum?
2. What is the significance of the second derivative in analyzing local minima?
3. How do you calculate the second derivative of a polynomial function?
4. What is the difference between local and global minimum values?
5. How would you approach finding critical points for a more complex function?

**Tip**: Always double-check critical points by evaluating both the first and second derivatives to confirm whether they are minima, maxima, or inflection points.

For 𝑘 ∈ ℝ, let 𝑓(𝑥) = 𝑥^4 + 2𝑥^3 + 𝑘𝑥^2 − 𝑘, 𝑥 ∈ ℝ. If 𝑥 = 3/2 is a point of local minima of 𝑓 and 𝑚 is the global minimum value of 𝑓, then 𝑓(0) − 𝑚 is equal to ____________ (in integer).

Explore the solution for finding the local and global minimum of the function f(x) = x^4 + 2x^3 + kx^2 − k, where x = 3/2 is a local minimum. This problem involves calculus concepts, including derivatives, local minima, and solving for k. Suitable for advanced students in Grades 11-12.

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