f(x) and g(x) are a differentiable function for all reals and h(x) = g[f(3x)]. The table below gives selected values for f(x), g(x), f '(x), and g '(x). Find the value of h '(1)

Numerical answers expected!

x 1 2 3 4 5 6
f(x) 0 3 2 1 2 0
g(x) 1 3 2 6 5 0
f'(x) 3 2 1 4 0 2
g'(x) 1 5 4 3 2 0

I got 15 for answer. Can you check for me, please?

To find the value of h'(1), we need to find the derivative of h(x) with respect to x and then substitute x = 1.

We have h(x) = g[f(3x)]. To find h'(x), we use the chain rule.

The chain rule states that if a function h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x).

Let's find h'(x) step by step:

1. Let's first find f'(x) and g'(x) using the information given in the table:

x | f(x) | g(x) | f'(x) | g'(x)
-----------------------------------------
1 | 0 | 1 | 3 | 1
2 | 3 | 3 | 2 | 5
3 | 2 | 2 | 1 | 4
4 | 1 | 6 | 4 | 3
5 | 2 | 5 | 0 | 2
6 | 0 | 0 | 2 | 0

2. Now, let's find f(3x) using the values of f(x):

f(3x) = [f(3 * 1), f(3 * 2), f(3 * 3), f(3 * 4), f(3 * 5), f(3 * 6)]
= [f(3), f(6), f(9), f(12), f(15), f(18)]
= [2, 0, 0, 2, 2, 2]

3. Now, let's find g[f(3x)] using the values of g(x) and f(3x):

g[f(3x)] = [g[f(3)], g[f(6)], g[f(9)], g[f(12)], g[f(15)], g[f(18)]]
= [g[2], g[0], g[0], g[2], g[2], g[2]]
= [2, 3, 3, 2, 2, 2]

4. Finally, let's find h'(x) using the values of f'(x), g'(x), and g[f(3x)]:

h'(x) = [g'[f(3)], g'[f(6)], g'[f(9)], g'[f(12)], g'[f(15)], g'[f(18)]]
= [g'[2], g'[0], g'[0], g'[2], g'[2], g'[2]]
= [4, 5, 5, 4, 4, 4]

5. Now, we can find h'(1) by substituting x = 1:

h'(1) = 4

Therefore, the value of h'(1) is 4.