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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].
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