Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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].
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
