Answer:
The number [tex]c = 3[/tex] satisfies the conclusion of Rolle's Theorem for [tex]f(x) = 5-24\cdot x +4\cdot x^{2}[/tex].
Step-by-step explanation:
According to the Rolle's Theorem, for all function continuous on [tex][a,b][/tex], there is a value [tex]c[/tex] ([tex]a \leq c\leq b[/tex]) such that:
[tex]f'(c) = \frac{f(b)-f(a)}{b-a}[/tex] (Eq. 1)
Where:
[tex]f'(c)[/tex] - First derivative of the function evaluated at [tex]x = c[/tex], dimensionless.
[tex]a[/tex], [tex]b[/tex] - Lower and upper bounds, dimensionless.
[tex]f(a)[/tex], [tex]f(b)[/tex] - Function evaluated at lower and upper bounds, dimensionless.
Let [tex]f(x) = 5-24\cdot x +4\cdot x^{2}[/tex], then upper and lower values are, respectively:
Lower bound ([tex]a = 2[/tex])
[tex]f(2) = 5-24\cdot (2) +4\cdot (2)^{2}[/tex]
[tex]f(2) = -27[/tex]
Upper bound ([tex]b = 4[/tex])
[tex]f(4) = 5-24\cdot (4) +4\cdot (4)^{2}[/tex]
[tex]f(4) = -27[/tex]
From Rolle's Theorem, we find that first derivative evaluated at [tex]c[/tex] is:
[tex]f'(c) = \frac{-27-(-27)}{4-2}[/tex]
[tex]f'(c) = 0[/tex]
Then, we find the first derivative of the function, equalize [tex]x[/tex] to [tex]c[/tex] and solve the resulting expression:
[tex]f'(x) = -24+8\cdot x[/tex]
[tex]-24+8\cdot c = 0[/tex]
[tex]c = 3[/tex]
The number [tex]c = 3[/tex] satisfies the conclusion of Rolle's Theorem for [tex]f(x) = 5-24\cdot x +4\cdot x^{2}[/tex].
