f(x) and g(x) are a differentiable function for all reals and h(x) = g[f(2x)]. The table below gives selected values for f(x), g(x), f '(x), and g '(x). Find the value of h '(1). (4 points)
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
16 btw
df/dx=2 not 3
To find the value of h'(1), we need to first find the derivative of h(x).
Using the chain rule, the derivative of h(x) = g[f(2x)] with respect to x is:
h'(x) = g'[f(2x)] * f'(2x) * 2
To find h'(1), we substitute x = 1 into the equation above:
h'(1) = g'[f(2)] * f'(2) * 2
To find the value of f(2), we look at the given table:
f(2) = 3
Similarly, g(3) = 2.
Using the table, we find the values of f'(2) and g'(3):
f'(2) = 2
g'(3) = 4
Substituting these values into the equation for h'(1):
h'(1) = g'[f(2)] * f'(2) * 2
= g'(3) * f'(2) * 2
= 4 * 2 * 2
= 16
Therefore, h'(1) = 16.
i don't know what the right answer is, but 24 was not considered the correct answer
noooooooo
h = g(f(2x))
dh/dx = dg/df * df/dx * 2
h'(1) = g'(f(2)) * 3 * 2
= g'(3)*6
= 4*6
= 24