Question

the table above gives values of f, f, g, and g at selected values of x. if h(x)=f(g(x)), then h(1)=

Explanation:

Step1: Recall the Chain Rule

The chain rule states that if \( h(x) = f(g(x)) \), then \( h'(x) = f'(g(x)) \cdot g'(x) \). So, to find \( h'(1) \), we need \( f'(g(1)) \cdot g'(1) \).

Step2: Find \( g(1) \) from the table

Looking at the table, when \( x = 1 \), \( g(x) = -1 \).

Step3: Find \( f'(g(1)) = f'(-1) \) from the table

Since \( g(1) = -1 \), we need \( f'(-1) \). From the table, when \( x = -1 \), \( f'(x) = 5 \).

Step4: Find \( g'(1) \) from the table

From the table, when \( x = 1 \), \( g'(x) = 2 \).

Step5: Apply the Chain Rule

Now, substitute into the chain rule formula: \( h'(1) = f'(g(1)) \cdot g'(1) = f'(-1) \cdot g'(1) = 5 \cdot 2 = 10 \)? Wait, no, wait. Wait, \( g(1) \): let's check the table again. The table for \( g(x) \): when \( x = -1 \), \( g(x) = 3 \); \( x = 1 \), \( g(x) = -1 \); \( x = 3 \), \( g(x) = 2 \)? Wait, no, the table columns: \( x \) has values -1, 1, 3. Then \( f(x) \): -1→6, 1→3, 3→1. \( f'(x) \): -1→5, 1→-3, 3→-2. \( g(x) \): -1→3, 1→-1, 3→2. \( g'(x) \): -1→-2, 1→2, 3→3. Oh! I made a mistake earlier. Let's correct:

\( g(1) \): when \( x = 1 \), \( g(x) = -1 \)? Wait, no, the \( g(x) \) row: for \( x = -1 \), \( g(x) = 3 \); \( x = 1 \), \( g(x) = -1 \); \( x = 3 \), \( g(x) = 2 \). Then \( g'(x) \): \( x = -1 \), \( g'(x) = -2 \); \( x = 1 \), \( g'(x) = 2 \); \( x = 3 \), \( g'(x) = 3 \). \( f'(x) \): \( x = -1 \), \( f'(x) = 5 \); \( x = 1 \), \( f'(x) = -3 \); \( x = 3 \), \( f'(x) = -2 \).

So, \( h(x) = f(g(x)) \), so \( h'(x) = f'(g(x)) \cdot g'(x) \).

So, \( h'(1) = f'(g(1)) \cdot g'(1) \).

First, find \( g(1) \): from \( g(x) \) row, \( x = 1 \), \( g(x) = -1 \).

Then, \( f'(g(1)) = f'(-1) \): from \( f'(x) \) row, \( x = -1 \), \( f'(x) = 5 \).

\( g'(1) \): from \( g'(x) \) row, \( x = 1 \), \( g'(x) = 2 \).

So, \( h'(1) = 5 \times 2 = 10 \)? Wait, but let's check again. Wait, maybe I mixed up \( g(x) \) values. Wait, the table:

\( x \)	\( f(x) \)	\( f'(x) \)	\( g(x) \)	\( g'(x) \)
1	3	-3	-1	2
3	1	-2	2	3

Ah! There we go. So \( g(1) = -1 \), correct. \( f'(-1) = 5 \), correct. \( g'(1) = 2 \), correct. So \( h'(1) = f'(-1) \times g'(1) = 5 \times 2 = 10 \). Wait, but the options have D as 10. So that's correct.

Wait, but earlier I thought maybe I messed up, but no. So the steps:

1. Chain rule: \( h'(x) = f'(g(x)) \cdot g'(x) \)
2. \( g(1) = -1 \) (from \( g(x) \) table at \( x=1 \))
3. \( f'(-1) = 5 \) (from \( f'(x) \) table at \( x=-1 \))
4. \( g'(1) = 2 \) (from \( g'(x) \) table at \( x=1 \))
5. Multiply: \( 5 \times 2 = 10 \)

Answer:

\( \boxed{10} \) (Wait, but the options have D as 10, so the answer is D. 10)