Answer:
Step-by-step explanation:
Given is a function
[tex]f(x)=-x^3+2x^2+2x+1[/tex]
[tex]f(6) = -6^3+2(6^2)+2(6)+1\\=-131\\[/tex]
[tex]f(-2) = -(-2)^3+2(-2)^2+2(-2)+1\\=13[/tex]
Average slope of this function is change of f(x) in (-2,6)/change of x in (-2,6)
= [tex]\frac{-131-13}{8} \\=-18[/tex]
By mean value theorem there exists a c such that f'(c) = -18
i.e. [tex]-3x^2+4x+2 =-8\\3x^2-4x-10 =0\\[/tex]
Using quadratic formula
x = 2.61, -1.277
Out of these only 2.61 lies in the given interval
c = 2.61