Answer:
(f∘g)(1) = 2; (f∘g)'(1) = 2
(f∘g)(2) = -2; (f∘g)'(2) = -2
Explanation:
You want (f∘g)(x) and (f∘g)'(x) for x=1 and x=2 given the function values and derivatives in the table.
(f∘g)(x)
This composition means f(g(x)). The value is found by first determining the value of z = g(x), then using that to find the value of f(z).
For x=1, the value of g(1) is seen to be -2.
For x=-2, the value of f(-2) is seen to be 2.
This means f(g(1)) = 2.
For x=2, the value of g(2) is 0.
For x= 0, the value of f(0) is -2.
This means f(g(2)) = -2.
(f∘g)'(x)
This is a little trickier, as you need to find the derivative of the composition:
f(g(x))' = f'(g(x))·g'(x)
In the attached table, we have made a column for f'(g(x)) to help find this product.
For x=1, f'(g(1)) = f'(-2) = 1; and g'(1) = 2, so f'(g(1))g'(1) = 1·2 = 2 = (f∘g)'(1)
For x=2, f'(g(2)) = f'(0) = 2; and g'(2) = -1, so f'(g(2))g'(2) = 2(-1) = -2 = (f∘g)'(2)