Answer :
To solve this problem, we'll follow a structured approach to find the mean slope of the function [tex]\( f(x) = 2x^3 - 3x^2 - 72x + 9 \)[/tex] on the interval [tex]\([-4, 8]\)[/tex] and identify the values of [tex]\( c \)[/tex] in this interval that satisfy the conditions of the Mean Value Theorem (MVT).
### Step 1: Calculate the Mean Slope
The mean slope of the function on the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Mean Slope} = \frac{f(b) - f(a)}{b - a} \][/tex]
For our function [tex]\( f(x) \)[/tex], we have [tex]\( a = -4 \)[/tex] and [tex]\( b = 8 \)[/tex]. We need to find [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex]:
1. Evaluate [tex]\( f(-4) \)[/tex]:
[tex]\[ f(-4) = 2(-4)^3 - 3(-4)^2 - 72(-4) + 9 = 2(-64) - 3(16) + 288 + 9 = -128 - 48 + 288 + 9 = 121 \][/tex]
2. Evaluate [tex]\( f(8) \)[/tex]:
[tex]\[ f(8) = 2(8)^3 - 3(8)^2 - 72(8) + 9 = 2(512) - 3(64) - 576 + 9 = 1024 - 192 - 576 + 9 = 265 \][/tex]
Now, compute the mean slope:
[tex]\[ \text{Mean Slope} = \frac{f(8) - f(-4)}{8 - (-4)} = \frac{265 - 121}{8 + 4} = \frac{144}{12} = 12 \][/tex]
### Step 2: Apply the Mean Value Theorem (MVT)
The MVT states that if [tex]\( f \)[/tex] is continuous on [tex]\([a, b]\)[/tex] and differentiable on [tex]\((a, b)\)[/tex], then there exists a [tex]\( c \)[/tex] in [tex]\((a, b)\)[/tex] such that:
[tex]\[ f'(c) = \text{Mean Slope} \][/tex]
### Step 3: Find [tex]\( f'(x) \)[/tex] and Solve for [tex]\( c \)[/tex]
First, find the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} (2x^3 - 3x^2 - 72x + 9) = 6x^2 - 6x - 72 \][/tex]
We need to find the values of [tex]\( c \)[/tex] in [tex]\((-4, 8)\)[/tex] such that [tex]\( f'(c) = 12 \)[/tex]:
Solve the equation:
[tex]\[ 6c^2 - 6c - 72 = 12 \][/tex]
First, simplify:
[tex]\[ 6c^2 - 6c - 72 - 12 = 0 \][/tex]
[tex]\[ 6c^2 - 6c - 84 = 0 \][/tex]
[tex]\[ c^2 - c - 14 = 0 \][/tex]
Solve the quadratic equation [tex]\( c^2 - c - 14 = 0 \)[/tex] using the quadratic formula [tex]\( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -14 \)[/tex]:
[tex]\[ c = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-14)}}{2(1)} = \frac{1 \pm \sqrt{1 + 56}}{2} = \frac{1 \pm \sqrt{57}}{2} \][/tex]
The solutions are:
[tex]\[ c = \frac{1 + \sqrt{57}}{2} \quad \text{and} \quad c = \frac{1 - \sqrt{57}}{2} \][/tex]
### Final Step: Determine Numerical Values
By evaluating expressions, we find:
The smaller value is:
[tex]\[ \frac{1 - \sqrt{57}}{2} \approx -3.2749 \][/tex]
The larger value is:
[tex]\[ \frac{1 + \sqrt{57}}{2} \approx 4.2749 \][/tex]
### Conclusion
The smaller value of [tex]\( c \)[/tex] is approximately [tex]\(-3.2749\)[/tex] and the larger value of [tex]\( c \)[/tex] is approximately [tex]\(4.2749\)[/tex]. Therefore,
The smaller one is [tex]\(-3.2749\)[/tex] and the larger one is [tex]\(4.2749\)[/tex].
### Step 1: Calculate the Mean Slope
The mean slope of the function on the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Mean Slope} = \frac{f(b) - f(a)}{b - a} \][/tex]
For our function [tex]\( f(x) \)[/tex], we have [tex]\( a = -4 \)[/tex] and [tex]\( b = 8 \)[/tex]. We need to find [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex]:
1. Evaluate [tex]\( f(-4) \)[/tex]:
[tex]\[ f(-4) = 2(-4)^3 - 3(-4)^2 - 72(-4) + 9 = 2(-64) - 3(16) + 288 + 9 = -128 - 48 + 288 + 9 = 121 \][/tex]
2. Evaluate [tex]\( f(8) \)[/tex]:
[tex]\[ f(8) = 2(8)^3 - 3(8)^2 - 72(8) + 9 = 2(512) - 3(64) - 576 + 9 = 1024 - 192 - 576 + 9 = 265 \][/tex]
Now, compute the mean slope:
[tex]\[ \text{Mean Slope} = \frac{f(8) - f(-4)}{8 - (-4)} = \frac{265 - 121}{8 + 4} = \frac{144}{12} = 12 \][/tex]
### Step 2: Apply the Mean Value Theorem (MVT)
The MVT states that if [tex]\( f \)[/tex] is continuous on [tex]\([a, b]\)[/tex] and differentiable on [tex]\((a, b)\)[/tex], then there exists a [tex]\( c \)[/tex] in [tex]\((a, b)\)[/tex] such that:
[tex]\[ f'(c) = \text{Mean Slope} \][/tex]
### Step 3: Find [tex]\( f'(x) \)[/tex] and Solve for [tex]\( c \)[/tex]
First, find the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} (2x^3 - 3x^2 - 72x + 9) = 6x^2 - 6x - 72 \][/tex]
We need to find the values of [tex]\( c \)[/tex] in [tex]\((-4, 8)\)[/tex] such that [tex]\( f'(c) = 12 \)[/tex]:
Solve the equation:
[tex]\[ 6c^2 - 6c - 72 = 12 \][/tex]
First, simplify:
[tex]\[ 6c^2 - 6c - 72 - 12 = 0 \][/tex]
[tex]\[ 6c^2 - 6c - 84 = 0 \][/tex]
[tex]\[ c^2 - c - 14 = 0 \][/tex]
Solve the quadratic equation [tex]\( c^2 - c - 14 = 0 \)[/tex] using the quadratic formula [tex]\( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -14 \)[/tex]:
[tex]\[ c = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-14)}}{2(1)} = \frac{1 \pm \sqrt{1 + 56}}{2} = \frac{1 \pm \sqrt{57}}{2} \][/tex]
The solutions are:
[tex]\[ c = \frac{1 + \sqrt{57}}{2} \quad \text{and} \quad c = \frac{1 - \sqrt{57}}{2} \][/tex]
### Final Step: Determine Numerical Values
By evaluating expressions, we find:
The smaller value is:
[tex]\[ \frac{1 - \sqrt{57}}{2} \approx -3.2749 \][/tex]
The larger value is:
[tex]\[ \frac{1 + \sqrt{57}}{2} \approx 4.2749 \][/tex]
### Conclusion
The smaller value of [tex]\( c \)[/tex] is approximately [tex]\(-3.2749\)[/tex] and the larger value of [tex]\( c \)[/tex] is approximately [tex]\(4.2749\)[/tex]. Therefore,
The smaller one is [tex]\(-3.2749\)[/tex] and the larger one is [tex]\(4.2749\)[/tex].
