https://file.aiquickdraw.com/mathbot/file/1730264299336l80d2l8a.jpg

To find the critical values of the function \( f(x) = 6x^3 + 42x^2 - 30x + 8 \), we need to follow these steps:

1. **Find the derivative** of \( f(x) \), denoted as \( f'(x) \).
2. **Set \( f'(x) = 0 \)** to locate points where the slope of the tangent line is zero, which may indicate critical points.
3. **Solve for \( x \)** to get the critical points.

Let's go through these steps.

### Step 1: Differentiate \( f(x) \)
The function is:
\[
f(x) = 6x^3 + 42x^2 - 30x + 8
\]
The derivative \( f'(x) \) is:
\[
f'(x) = 18x^2 + 84x - 30
\]

### Step 2: Set \( f'(x) = 0 \)
Now, set the derivative equal to zero:
\[
18x^2 + 84x - 30 = 0
\]

### Step 3: Simplify the Equation
Divide every term by 6 to simplify:
\[
3x^2 + 14x - 5 = 0
\]

### Step 4: Solve the Quadratic Equation
To solve \( 3x^2 + 14x - 5 = 0 \), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 3 \), \( b = 14 \), and \( c = -5 \).

Substitute the values:
\[
x = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3}
\]
\[
x = \frac{-14 \pm \sqrt{196 + 60}}{6}
\]
\[
x = \frac{-14 \pm \sqrt{256}}{6}
\]
\[
x = \frac{-14 \pm 16}{6}
\]

This gives two solutions:
1. \( x = \frac{-14 + 16}{6} = \frac{2}{6} = \frac{1}{3} \)
2. \( x = \frac{-14 - 16}{6} = \frac{-30}{6} = -5 \)

### Conclusion
The critical values are:
\[
x = \frac{1}{3}, -5
\]

So, the critical values are \( \frac{1}{3} \) and \( -5 \).

Would you like more details or have any questions?

Here are some related questions for further understanding:

1. How do critical points relate to the shape of the graph of a function?
2. What does the sign of the second derivative tell us about the critical points?
3. Can there be critical points where the derivative does not exist?
4. How do we classify critical points as minima, maxima, or saddle points?
5. How does the degree of a polynomial affect the number of potential critical points?

**Tip:** Always check for points where the derivative does not exist, as they can also be critical points.

Determine the critical values of f(x) = 6x^3 + 42x^2 - 30x + 8. Separate your answers with commas. Leave answers as integers or reduced fractions.

Learn how to find the critical values of the polynomial function f(x) = 6x^3 + 42x^2 - 30x + 8. This step-by-step solution involves taking the derivative, setting it equal to zero, and using the quadratic formula to solve. Ideal for high school calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation