Find the absolute maximum and absolute minimum values of the functionf(x)=x3−6x2−63x+5over each of the indicated intervals.

(a) Interval = [−4,0].

**1.**  

Absolute maximum

=

**2.**  

Absolute minimum

=

(b) Interval = [−1,8].

**1.**  

Absolute maximum

=

**2.**  

Absolute minimum

=

(c) Interval = [−4,8].

**1.**  

Absolute maximum

=

**2.**  

Absolute minimum

=

To find the absolute maximum and absolute minimum values of the function \( f(x) = x^3 - 6x^2 - 63x + 5 \) over each of the given intervals, we need to follow these steps:

1. Find the critical points of \( f(x) \) by setting its derivative \( f'(x) \) equal to zero.
2. Evaluate \( f(x) \) at the critical points and the endpoints of the given intervals.
3. Compare these values to determine the absolute maximum and minimum values over each interval.

### Step 1: Find the critical points

First, we find the derivative of \( f(x) \):
\[ f'(x) = 3x^2 - 12x - 63 \]

Set the derivative equal to zero to find the critical points:
\[ 3x^2 - 12x - 63 = 0 \]

Divide the equation by 3:
\[ x^2 - 4x - 21 = 0 \]

Solve the quadratic equation:
\[ x = \frac{4 \pm \sqrt{16 + 84}}{2} \]
\[ x = \frac{4 \pm \sqrt{100}}{2} \]
\[ x = \frac{4 \pm 10}{2} \]

So, the critical points are:
\[ x = 7 \]
\[ x = -3 \]

### Step 2: Evaluate \( f(x) \) at the critical points and endpoints of each interval

**(a) Interval = \([-4, 0]\)**

Endpoints: \( x = -4 \) and \( x = 0 \)
Critical points within the interval: \( x = -3 \)

Calculate \( f(x) \) at these points:
\[ f(-4) = (-4)^3 - 6(-4)^2 - 63(-4) + 5 = -64 - 96 + 252 + 5 = 97 \]
\[ f(0) = 0^3 - 6(0)^2 - 63(0) + 5 = 5 \]
\[ f(-3) = (-3)^3 - 6(-3)^2 - 63(-3) + 5 = -27 - 54 + 189 + 5 = 113 \]

**1.** Absolute maximum = 113 (at \( x = -3 \))
**2.** Absolute minimum = 5 (at \( x = 0 \))

**(b) Interval = \([-1, 8]\)**

Endpoints: \( x = -1 \) and \( x = 8 \)
Critical points within the interval: \( x = 7 \)

Calculate \( f(x) \) at these points:
\[ f(-1) = (-1)^3 - 6(-1)^2 - 63(-1) + 5 = -1 - 6 + 63 + 5 = 61 \]
\[ f(8) = 8^3 - 6(8)^2 - 63(8) + 5 = 512 - 384 - 504 + 5 = -371 \]
\[ f(7) = 7^3 - 6(7)^2 - 63(7) + 5 = 343 - 294 - 441 + 5 = -387 \]

**1.** Absolute maximum = 61 (at \( x = -1 \))
**2.** Absolute minimum = -387 (at \( x = 7 \))

**(c) Interval = \([-4, 8]\)**

Endpoints: \( x = -4 \) and \( x = 8 \)
Critical points within the interval: \( x = -3 \) and \( x = 7 \)

Calculate \( f(x) \) at these points:
\[ f(-4) = (-4)^3 - 6(-4)^2 - 63(-4) + 5 = 97 \]
\[ f(8) = 8^3 - 6(8)^2 - 63(8) + 5 = -371 \]
\[ f(-3) = (-3)^3 - 6(-3)^2 - 63(-3) + 5 = 113 \]
\[ f(7) = 7^3 - 6(7)^2 - 63(7) + 5 = -387 \]

**1.** Absolute maximum = 113 (at \( x = -3 \))
**2.** Absolute minimum = -387 (at \( x = 7 \))

### Summary:

(a) Interval = \([-4, 0]\)
**1.** Absolute maximum = 113 (at \( x = -3 \))
**2.** Absolute minimum = 5 (at \( x = 0 \))

(b) Interval = \([-1, 8]\)
**1.** Absolute maximum = 61 (at \( x = -1 \))
**2.** Absolute minimum = -387 (at \( x = 7 \))

(c) Interval = \([-4, 8]\)
**1.** Absolute maximum = 113 (at \( x = -3 \))
**2.** Absolute minimum = -387 (at \( x = 7 \))

Would you like more details or have any questions?

### Suggested Questions:
1. How do you find the derivative of a function?
2. What are critical points and why are they important?
3. How do you solve a quadratic equation?
4. Can you explain how to evaluate a function at a given point?
5. What is the difference between an absolute maximum and a relative maximum?

### Tip:
When finding the maximum and minimum values of a function on a closed interval, always check both the critical points within the interval and the values at the endpoints.

Discover how to find the absolute maximum and minimum values of the function f(x) = x^3 - 6x^2 - 63x + 5 over specified intervals [-4, 0], [-1, 8], and [-4, 8]. This problem involves calculus concepts like critical points and derivatives of cubic functions. Suitable for advanced high school students.

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