Answer:
Yes , it satisfies the hypothesis and we can conclude that c = 1 is an element of [0,2]
c = 1 ∈ [0,2]
Step-by-step explanation:
Given that:
[tex]f(x) = 4x^2 -3x + 2, [0, 2][/tex]
which is read as the function of x = 4x² - 3x + 2 along the interval [0,2]
Differentiating the function with respect to x is;
f(x) = 8x - 3
Using the Mean value theorem to see if the function satisfies it, we have:
[tex]f'c = \dfrac{f(b)-f(a)}{b-a}[/tex]
[tex]8c -3 = \dfrac{f(2)-f(0)}{2-0}[/tex]
since the polynomial function is differentiated in [0,2]
[tex]8c -3 = \dfrac{(4(2)^2-3(2)+2)-(4(0)^2-3(0)+2)}{2-0}[/tex]
[tex]8c -3 = \dfrac{(4(4)-3(2)+2)-(4(0)-3(0)+2)}{2-0}[/tex]
[tex]8c -3 = \dfrac{(16-6+2)-(0-0+2)}{2-0}[/tex]
[tex]8c -3 = \dfrac{(12)-(2)}{2}[/tex]
[tex]8c -3 = \dfrac{10}{2}[/tex]
8c -3 = 5
8c = 5+3
8c = 8
c = 8/8
c = 1
Therefore, we can conclude that c = 1 is an element of [0,2]
c = 1 ∈ [0,2]
