Question 1027573
By the mean value theorem,

there is a {{{0<= alpha <= x}}} such that {{{f(x) = f(0) + (df(alpha)/dx)x}}}

==> {{{f(x) <= f(0) + 6x = 2 = 6x}}}  because  {{{df(x)/dx <= 6}}} for all x in [-10,10].

A)  ==> {{{f(4) <= 2 + 6*4 = 26}}} ==> maximum possible value of f(4) is 26.

B)  ==> {{{f(2) <= 2 + 6*2 = 14}}} ==> maximum possible value of f(2) is 14.

C)  Can f(5) be negative? YES. The derivative in the neighborhood around x = 5 can be negative enough (and hence the graph fall drastically enough) so as to warrant the graph of of f(x) to cross the x-axis before x = 5.  (Remember, f'(x) is less than or equal to 6 in [-10,10].)

Can f(5) = 0? YES, by a reasoning very similar to the preceding paragraph.

D) Can f(5) = 31? YES.
{{{f(5) <= 2 + 6*5 = 32}}} ==> maximum possible value of f(5) is 32.

Can f(5) = 35?  NO. The highest f(5) can get is 32