Answer :
To solve the problem of finding the critical points and evaluating the function [tex]\(f(x) = x^3 - x^2 - 6x + 2\)[/tex] on the interval [tex]\([0, 3]\)[/tex], follow these steps:
1. Find the first derivative of the function:
Given the function [tex]\(f(x) = x^3 - x^2 - 6x + 2\)[/tex], we need to find its first derivative [tex]\(f'(x)\)[/tex]. The first derivative is calculated as follows:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - x^2 - 6x + 2) = 3x^2 - 2x - 6 \][/tex]
2. Find the critical points:
Critical points occur where the first derivative [tex]\(f'(x)\)[/tex] is equal to zero:
[tex]\[ 3x^2 - 2x - 6 = 0 \][/tex]
Solving this quadratic equation gives the critical points:
[tex]\[ x = \frac{1}{3} - \frac{\sqrt{19}}{3}, \quad x = \frac{1}{3} + \frac{\sqrt{19}}{3} \][/tex]
3. Determine which critical points lie within the interval [tex]\([0, 3]\)[/tex]:
We need to check if the critical points lie within the given interval. Evaluating the critical points:
- [tex]\(\frac{1}{3} - \frac{\sqrt{19}}{3}\)[/tex] is less than 0 (so it is outside the interval).
- [tex]\(\frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex] is inside the interval [tex]\([0, 3]\)[/tex].
Therefore, the only critical point within the interval is:
[tex]\[ x = \frac{1}{3} + \frac{\sqrt{19}}{3} \][/tex]
4. Evaluate the function at the endpoints of the interval and at the critical points within the interval:
The endpoints of the interval are 0 and 3. We also have the critical point [tex]\(\frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex]. We evaluate [tex]\(f(x)\)[/tex] at these points:
- Evaluate at [tex]\(x = 0\)[/tex]:
[tex]\[ f(0) = (0)^3 - (0)^2 - 6(0) + 2 = 2 \][/tex]
- Evaluate at [tex]\(x = 3\)[/tex]:
[tex]\[ f(3) = (3)^3 - (3)^2 - 6(3) + 2 = 27 - 9 - 18 + 2 = 2 \][/tex]
- Evaluate at [tex]\(x = \frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex]:
[tex]\[ f\left(\frac{1}{3} + \frac{\sqrt{19}}{3}\right) \approx -6.20882073535354 \][/tex]
5. Summarize the results:
The first derivative is [tex]\(3x^2 - 2x - 6\)[/tex].
The critical points are [tex]\(\frac{1}{3} - \frac{\sqrt{19}}{3}\)[/tex] and [tex]\(\frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex], with the latter being within the given interval.
The points evaluated are [tex]\(\frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex], 0, and 3.
The corresponding function values are:
- At [tex]\(x = \frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex] the value is approximately [tex]\(-6.20882073535354\)[/tex].
- At [tex]\(x = 0\)[/tex] the value is [tex]\(2\)[/tex].
- At [tex]\(x = 3\)[/tex] the value is [tex]\(2\)[/tex].
By analyzing these values, you can determine the behavior of the function over the given interval.
1. Find the first derivative of the function:
Given the function [tex]\(f(x) = x^3 - x^2 - 6x + 2\)[/tex], we need to find its first derivative [tex]\(f'(x)\)[/tex]. The first derivative is calculated as follows:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - x^2 - 6x + 2) = 3x^2 - 2x - 6 \][/tex]
2. Find the critical points:
Critical points occur where the first derivative [tex]\(f'(x)\)[/tex] is equal to zero:
[tex]\[ 3x^2 - 2x - 6 = 0 \][/tex]
Solving this quadratic equation gives the critical points:
[tex]\[ x = \frac{1}{3} - \frac{\sqrt{19}}{3}, \quad x = \frac{1}{3} + \frac{\sqrt{19}}{3} \][/tex]
3. Determine which critical points lie within the interval [tex]\([0, 3]\)[/tex]:
We need to check if the critical points lie within the given interval. Evaluating the critical points:
- [tex]\(\frac{1}{3} - \frac{\sqrt{19}}{3}\)[/tex] is less than 0 (so it is outside the interval).
- [tex]\(\frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex] is inside the interval [tex]\([0, 3]\)[/tex].
Therefore, the only critical point within the interval is:
[tex]\[ x = \frac{1}{3} + \frac{\sqrt{19}}{3} \][/tex]
4. Evaluate the function at the endpoints of the interval and at the critical points within the interval:
The endpoints of the interval are 0 and 3. We also have the critical point [tex]\(\frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex]. We evaluate [tex]\(f(x)\)[/tex] at these points:
- Evaluate at [tex]\(x = 0\)[/tex]:
[tex]\[ f(0) = (0)^3 - (0)^2 - 6(0) + 2 = 2 \][/tex]
- Evaluate at [tex]\(x = 3\)[/tex]:
[tex]\[ f(3) = (3)^3 - (3)^2 - 6(3) + 2 = 27 - 9 - 18 + 2 = 2 \][/tex]
- Evaluate at [tex]\(x = \frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex]:
[tex]\[ f\left(\frac{1}{3} + \frac{\sqrt{19}}{3}\right) \approx -6.20882073535354 \][/tex]
5. Summarize the results:
The first derivative is [tex]\(3x^2 - 2x - 6\)[/tex].
The critical points are [tex]\(\frac{1}{3} - \frac{\sqrt{19}}{3}\)[/tex] and [tex]\(\frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex], with the latter being within the given interval.
The points evaluated are [tex]\(\frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex], 0, and 3.
The corresponding function values are:
- At [tex]\(x = \frac{1}{3} + \frac{\sqrt{19}}{3}\)[/tex] the value is approximately [tex]\(-6.20882073535354\)[/tex].
- At [tex]\(x = 0\)[/tex] the value is [tex]\(2\)[/tex].
- At [tex]\(x = 3\)[/tex] the value is [tex]\(2\)[/tex].
By analyzing these values, you can determine the behavior of the function over the given interval.
