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 $f'(x)$:

= 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.