Answer:
[tex]c = 5[/tex]
Step-by-step explanation:
Given
[tex]f(c) = M[/tex]
[tex]f(x) = x^2 - x + 1[/tex]
[tex]Interval: [1,8][/tex]
[tex]M = 21[/tex]
Required
Find c using Intermediate Value theorem
First, check if the value of M is within the given range:
[tex]f(x) = x^2 - x + 1[/tex]
[tex]f(1) = 1^2 - 1 + 1[/tex]
[tex]f(1) = 1[/tex]
[tex]f(x) = 8^2 - 8 + 1[/tex]
[tex]f(x) = 57[/tex]
[tex]1 \le M \le 57[/tex]
[tex]1 \le 21 \le 57[/tex]
M is within range.
Solving further:
We have:
[tex]f(c) = f(x) = M[/tex]
[tex]f(x) = 21[/tex]
Substitute 21 for f(x) in [tex]f(x) = x^2 - x + 1[/tex]
[tex]21 = x^2 - x + 1[/tex]
Express as quadratic function
[tex]x^2 - x + 1 - 21 = 0[/tex]
[tex]x^2 - x - 20 = 0[/tex]
Expand
[tex]x^2 + 4x - 5x - 20[/tex]
[tex]x(x+4)-5(x+4)=0[/tex]
[tex](x - 5)(x+4) = 0[/tex]
[tex]x - 5 = 0[/tex] or [tex]x + 4= 0[/tex]
[tex]x = 5[/tex] or [tex]x = -4[/tex]
The value of [tex]x = -4[/tex] is outside the [tex]Interval: [1,8][/tex]
So:
[tex]x = 5[/tex]
[tex]f(c) = f(x) = M[/tex]
[tex]f(c) = f(5) = 21[/tex]
By comparison:
[tex]c = 5[/tex]
