Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To find the value of [tex]\( c \)[/tex] such that [tex]\( f(c) = M \)[/tex] for the function [tex]\( f(x) = x^2 - x + 1 \)[/tex] on the interval [tex]\([1, 11]\)[/tex] where [tex]\( M = 31 \)[/tex], we can follow these steps:
1. Review the function and interval: We are given the function [tex]\( f(x) = x^2 - x + 1 \)[/tex] and the interval [tex]\( [1, 11] \)[/tex].
2. Apply the Intermediate Value Theorem: The Intermediate Value Theorem states that if a continuous function [tex]\( f \)[/tex] takes on two values [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex] over an interval [tex]\([a, b]\)[/tex], then it must take on any value between [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex] within that interval.
3. Evaluate the function at the endpoints:
- Calculate [tex]\( f(1) \)[/tex]:
[tex]\[ f(1) = 1^2 - 1 + 1 = 1 \][/tex]
- Calculate [tex]\( f(11) \)[/tex]:
[tex]\[ f(11) = 11^2 - 11 + 1 = 121 - 11 + 1 = 111 \][/tex]
We can now see that [tex]\( f(x) \)[/tex] takes on values from 1 to 111 within the interval [tex]\([1, 11]\)[/tex].
4. Check if [tex]\( M = 31 \)[/tex] lies between [tex]\( f(1) \)[/tex] and [tex]\( f(11) \)[/tex]: Since [tex]\( 1 \)[/tex] and [tex]\( 111 \)[/tex] bracket [tex]\( 31 \)[/tex] (i.e., [tex]\( 1 < 31 < 111 \)[/tex]), by the Intermediate Value Theorem, there must be some [tex]\( c \)[/tex] in [tex]\([1, 11]\)[/tex] such that [tex]\( f(c) = 31 \)[/tex].
5. Find the specific value of [tex]\( c \)[/tex]:
To find the specific value of [tex]\( c \)[/tex] for which [tex]\( f(c) = 31 \)[/tex]:
Set the equation:
[tex]\[ c^2 - c + 1 = 31 \][/tex]
Rearrange to a standard quadratic equation form:
[tex]\[ c^2 - c + 1 - 31 = 0 \][/tex]
[tex]\[ c^2 - c - 30 = 0 \][/tex]
6. Solve the quadratic equation:
Use the quadratic formula [tex]\( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
For [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -30 \)[/tex]:
[tex]\[ c = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \][/tex]
[tex]\[ c = \frac{1 \pm \sqrt{1 + 120}}{2} \][/tex]
[tex]\[ c = \frac{1 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ c = \frac{1 \pm 11}{2} \][/tex]
This gives us two solutions:
[tex]\[ c = \frac{1 + 11}{2} = \frac{12}{2} = 6 \][/tex]
[tex]\[ c = \frac{1 - 11}{2} = \frac{-10}{2} = -5 \][/tex]
7. Check the solution within the interval:
Since [tex]\( c = -5 \)[/tex] is not within the interval [tex]\([1, 11]\)[/tex], the valid solution for [tex]\( c \)[/tex] is:
[tex]\[ c = 6 \][/tex]
Therefore, the value of [tex]\( c \)[/tex] such that [tex]\( f(c) = M \)[/tex] is [tex]\( \boxed{6} \)[/tex].
1. Review the function and interval: We are given the function [tex]\( f(x) = x^2 - x + 1 \)[/tex] and the interval [tex]\( [1, 11] \)[/tex].
2. Apply the Intermediate Value Theorem: The Intermediate Value Theorem states that if a continuous function [tex]\( f \)[/tex] takes on two values [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex] over an interval [tex]\([a, b]\)[/tex], then it must take on any value between [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex] within that interval.
3. Evaluate the function at the endpoints:
- Calculate [tex]\( f(1) \)[/tex]:
[tex]\[ f(1) = 1^2 - 1 + 1 = 1 \][/tex]
- Calculate [tex]\( f(11) \)[/tex]:
[tex]\[ f(11) = 11^2 - 11 + 1 = 121 - 11 + 1 = 111 \][/tex]
We can now see that [tex]\( f(x) \)[/tex] takes on values from 1 to 111 within the interval [tex]\([1, 11]\)[/tex].
4. Check if [tex]\( M = 31 \)[/tex] lies between [tex]\( f(1) \)[/tex] and [tex]\( f(11) \)[/tex]: Since [tex]\( 1 \)[/tex] and [tex]\( 111 \)[/tex] bracket [tex]\( 31 \)[/tex] (i.e., [tex]\( 1 < 31 < 111 \)[/tex]), by the Intermediate Value Theorem, there must be some [tex]\( c \)[/tex] in [tex]\([1, 11]\)[/tex] such that [tex]\( f(c) = 31 \)[/tex].
5. Find the specific value of [tex]\( c \)[/tex]:
To find the specific value of [tex]\( c \)[/tex] for which [tex]\( f(c) = 31 \)[/tex]:
Set the equation:
[tex]\[ c^2 - c + 1 = 31 \][/tex]
Rearrange to a standard quadratic equation form:
[tex]\[ c^2 - c + 1 - 31 = 0 \][/tex]
[tex]\[ c^2 - c - 30 = 0 \][/tex]
6. Solve the quadratic equation:
Use the quadratic formula [tex]\( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
For [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -30 \)[/tex]:
[tex]\[ c = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \][/tex]
[tex]\[ c = \frac{1 \pm \sqrt{1 + 120}}{2} \][/tex]
[tex]\[ c = \frac{1 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ c = \frac{1 \pm 11}{2} \][/tex]
This gives us two solutions:
[tex]\[ c = \frac{1 + 11}{2} = \frac{12}{2} = 6 \][/tex]
[tex]\[ c = \frac{1 - 11}{2} = \frac{-10}{2} = -5 \][/tex]
7. Check the solution within the interval:
Since [tex]\( c = -5 \)[/tex] is not within the interval [tex]\([1, 11]\)[/tex], the valid solution for [tex]\( c \)[/tex] is:
[tex]\[ c = 6 \][/tex]
Therefore, the value of [tex]\( c \)[/tex] such that [tex]\( f(c) = M \)[/tex] is [tex]\( \boxed{6} \)[/tex].
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. 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.
