Consider the function f(x)=3/4x^4-x^3-9x^2+20 (a) Find f(-2) (b) Find f'(x) The graph of the function f has a local minimum at the point where x=-2 (c) Using your answer to part (b), show that there is a second local minimum at x = 3

Let’s solve the parts step-by-step:

### **(a) Find \(f(-2)\)**  
Given the function:  
\[
f(x) = \frac{3}{4}x^4 - x^3 - 9x^2 + 20
\]
Substitute \(x = -2\) into the function:

\[
f(-2) = \frac{3}{4}(-2)^4 - (-2)^3 - 9(-2)^2 + 20
\]

Simplify each term:
\[
(-2)^4 = 16 \quad \text{so} \quad \frac{3}{4}(16) = 12
\]
\[
(-2)^3 = -8
\]
\[
(-2)^2 = 4 \quad \text{so} \quad 9(4) = 36
\]

Now plug everything into the expression:
\[
f(-2) = 12 - (-8) - 36 + 20
\]
\[
f(-2) = 12 + 8 - 36 + 20 = 4
\]

Thus,  
\[
f(-2) = 4
\]

---

### **(b) Find \(f'(x)\)**  
We differentiate \(f(x) = \frac{3}{4}x^4 - x^3 - 9x^2 + 20\).

Apply the power rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\).

\[
f'(x) = \frac{3}{4} \cdot 4x^3 - 3x^2 - 18x + 0
\]

Simplify:
\[
f'(x) = 3x^3 - 3x^2 - 18x
\]

Thus,  
\[
f'(x) = 3x^3 - 3x^2 - 18x
\]

---

### **(c) Show that there is a second local minimum at \(x = 3\)**  
To check for local minima, we need to find the critical points by setting \(f'(x) = 0\).

\[
3x^3 - 3x^2 - 18x = 0
\]

Factor out \(3x\):

\[
3x(x^2 - x - 6) = 0
\]

Solve the quadratic inside the parentheses:

\[
x^2 - x - 6 = 0
\]

Factor:

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

So, the solutions are:

\[
x = 3 \quad \text{and} \quad x = -2
\]

These are the critical points. Since we are given that there is a local minimum at \(x = -2\), we now check whether \(x = 3\) also corresponds to a local minimum. 

---

### **Second Derivative Test**

We use the second derivative to determine if the function has a local minimum at \(x = 3\). 

Start by finding \(f''(x)\).  
Differentiate \(f'(x) = 3x^3 - 3x^2 - 18x\).

\[
f''(x) = 9x^2 - 6x - 18
\]

Now evaluate \(f''(x)\) at \(x = 3\):

\[
f''(3) = 9(3)^2 - 6(3) - 18
\]

Simplify:

\[
f''(3) = 9(9) - 18 - 18 = 81 - 36 = 45
\]

Since \(f''(3) = 45 > 0\), the function is concave up at \(x = 3\). Therefore, there is a **local minimum** at \(x = 3\).

---

### **Final Answer Summary**
- **(a)** \(f(-2) = 4\)
- **(b)** \(f'(x) = 3x^3 - 3x^2 - 18x\)
- **(c)** There is a second local minimum at \(x = 3\) since \(f''(3) = 45 > 0\), indicating concave up.

---

Do you have any questions or need further details?  
Here are five related questions to explore:
1. What is the value of \(f(3)\)?
2. How would the function behave as \(x \to \infty\) or \(x \to -\infty\)?
3. Can you graph the function to visualize the local minima and maxima?
4. How do you find points of inflection for the function?
5. What are the x-intercepts of the original function \(f(x)\)?

**Tip:** Always verify the sign of the second derivative at critical points to determine if it’s a local minimum or maximum.

Learn how to find the local minima of the function f(x) = 3/4x^4 - x^3 - 9x^2 + 20. This problem covers key calculus concepts such as differentiation, critical points, and the second derivative test, suitable for Grades 11-12. Step-by-step solution included.

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