1. Part (a)
h(1) = f(g(1)) - 6
h(3) = f(g(3)) - 6
h(1) = f(2) - 6
h(3) = f(4) - 6
h(1) = 9 - 6
h(3) =
h(1) = 3
h(3) = -7
Since f(x) & g(x) are differentiable functions, they must also be continuous. Therefore, h(x) is continuous as well.
If h(1) = 3 and h(3) = -7, then the Intermediate Value Theorem states that h(x) must be -5 somewhere between x=1 & x=3.

2. Part (b)
From Part (a), we’ve seen that h(x) passes through the points (1,3) and (3,-7).
If we connect these two points with a line and find its slope, it will be m=-5.
The Mean Value Theorem (MVT) states that there must be at least one value “c” on the interval 1 < c < 3 such that h’(c) = -5

3. This is the Fundamental Theorem of Calculus (FTC) !!!
Part (c)
This is the Fundamental Theorem of Calculus (FTC) !!!
w’(3) = f(g(3)) * g’(3) - f(1) * 0
w’(3) = f(4) * 2
w’(3) = -1 * 2
w’(3) = -2

4. Therefore, the slope of g-1(x) @ (2,1) is 1/5.
Part (d)
Therefore, the slope of (2,1) is 1/5.
The slope of (1,2) was 5/1.
g(x) passes through the following points:
(1,2)
(2,3)
(3,4)
(4,6)
Therefore, g-1(x) will pass through these points:
(2,1)
(3,2)
(4,3)
(6,4)
(y-1) = m (x-2)
1/5 or
y = 1/5 x + 3/5