A) First, figure out what h(2) and h(8) are
h(2) = f(g(2)) - (2)^2 = f(6) - 4 = 9 - 4 = 5
h(8) = f(g(8)) - (8)^2 = f(4) - 64 = 12 - 64 = -52
Since f and g are both continuous at all values of x, so is h. Because h is continuous and must pass through (2,5) and (8,-52), there must be some value 2 < c < 8 where h(c) = 0, due to the intermediate value theorem
B) You didn't specify which function we're trying to find the tangent lines for, so I'll just tell you how to figure it out:
First, get the value of the first derivative of the function at x = 2 and x = 8. You have a table, so just use f'(x) or g'(x)
As you should know, this value is the slope of the corresponding non-derivative function at the point specified. If the slopes from both x = 2 and x = 8 are not the same, the two tangent lines are not parallel.
C) Similar to part A, the first step is to find the value of h(x) at x = 4 and x = 6. The reason we're finding the value of h(x) instead of h'(x) is because of the mean value theorem, which we'll soon put to work.
h(4) = f(g(4)) - (4)^2 = f(8) - 16 = 18 - 16 = 2
h(6) = f(g(6)) - (6)^2 = f(2) - 36 = 38 - 36 = 2
Since you know the function h(x) must pass through (4,2) and (6,2), you know that there must be a point 4 < c < 6 where h'(c) has to equal 0 because of the aforementioned mean value theorem. In layman's terms, if the average slope of a function between two points is 0 and the function is continuous and differentiable at all values of x, there is no possible function line that can be drawn between the two where the slope doesn't hit 0 at some point. This holds true for all real slope values too, not just 0
