https://file.aiquickdraw.com/mathbot/file/1729896165990loaw5qoi.jpg

The given problem involves the function \( f(x) = \frac{1}{2x^2 - 3x + 6} \) and requires the following:

### **(a)**  
Using MATLAB to graph \( f(x) \) on the interval \([-3, 3]\), and zooming in to find the point where \( f(x) \) is maximized. This part is software-specific, so I will focus on how you can do this:

- **Steps to graph in MATLAB**:
    ```matlab
    f = @(x) 1 ./ (2*x.^2 - 3*x + 6);
    fplot(f, [-3, 3])
    grid on
    ```
    - Use MATLAB's interactive tools to zoom in and read off the maximum value from the plot.

---

### **(b)**  
**Find the derivative of \( f(x) \).**

Using the quotient rule:
\[
\frac{d}{dx} \left( \frac{1}{2x^2 - 3x + 6} \right) = \frac{-\left( 2x^2 - 3x + 6 \right)'}{(2x^2 - 3x + 6)^2}
\]
Now, differentiate the numerator \( 2x^2 - 3x + 6 \):
\[
(2x^2 - 3x + 6)' = 4x - 3
\]
So the derivative is:
\[
f'(x) = \frac{-(4x - 3)}{(2x^2 - 3x + 6)^2}
\]

---

### **(c)**  
**Find the exact value of \( x \) where \( f(x) \) reaches its maximum.**

To find the critical points, set the derivative \( f'(x) \) equal to zero:
\[
\frac{-(4x - 3)}{(2x^2 - 3x + 6)^2} = 0
\]
The numerator must be zero, so:
\[
4x - 3 = 0 \implies x = \frac{3}{4}
\]

To confirm if this is a maximum, you can take the second derivative (or analyze the behavior of the first derivative around this point). Evaluating the function at \( x = \frac{3}{4} \):
\[
f\left( \frac{3}{4} \right) = \frac{1}{2\left( \frac{3}{4} \right)^2 - 3\left( \frac{3}{4} \right) + 6}
\]
After simplifying:
\[
f\left( \frac{3}{4} \right) = \frac{1}{2 \times \frac{9}{16} - \frac{9}{4} + 6} = \frac{1}{\frac{9}{8} - \frac{18}{8} + \frac{48}{8}} = \frac{1}{\frac{39}{8}} = \frac{8}{39}
\]

So the function reaches its maximum value at \( x = \frac{3}{4} \).

---

### **(d)**  
**At what point is the graph of \( f(x) \) rising most steeply?**

The steepest increase in the function corresponds to the point where the derivative \( f'(x) \) is maximized. To do this, we need to find where the absolute value of \( f'(x) \) reaches a peak.

Graph the derivative function \( f'(x) = \frac{-(4x - 3)}{(2x^2 - 3x + 6)^2} \) in MATLAB using:
```matlab
f_prime = @(x) -(4*x - 3) ./ (2*x.^2 - 3*x + 6).^2;
fplot(f_prime, [-3, 3])
grid on
```

Look for the point where \( f'(x) \) achieves its highest value. This might require numerical methods or a close look at the graph.

---

Do you want a detailed solution of any part or assistance with MATLAB code?  

---

Here are **five related questions** to deepen your understanding:
1. How do you apply the quotient rule for more complex rational functions?
2. How do second derivatives help confirm local maxima or minima?
3. What happens to \( f(x) \) as \( x \to \infty \)?
4. Why does the maximum value occur where the derivative is zero?
5. How can we interpret the graph of \( f'(x) \) to find points of steepest change?

**Tip:** Always verify maxima and minima by checking the second derivative or analyzing the behavior of \( f'(x) \) around critical points.

Let f(x) = 1 / (2x^2 - 3x + 6). Solve the following: (a) Using MATLAB, graph f(x) on the interval [-3,3]. By zooming in, find the point where f(x) is maximized to two decimal place accuracy. (b) Find the derivative of f(x). (c) Using part (b), find the exact value of x where f(x) reaches its maximum value. (d) At what point is the graph of f(x) rising most steeply? Describe how you determined the location of this point.

Explore how to graph and analyze the function f(x) = 1 / (2x^2 - 3x + 6), find its derivative, and locate the points of maximum value and steepest rise. This problem covers important calculus concepts like the quotient rule and finding critical points.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation