Suppose the functions f and g and their derivatives have the following values at x = 1 and x = 2. Let h(x) = f(g(x)). Evaluate h′(1).
X | f(x) G(x) f'(x) g'(x)
____________________________
1 | 8 2 1/3 -3
2 | 3 -4 2π 5
My answer: -6π but I'm not really sure if I did this right
g(1) = 2
f'(2) = 2π
h'(1) = f'(2)*g'(1) = -6π
you are correct
To evaluate h'(1), we need to find the derivative of h(x) with respect to x and then evaluate it at x = 1.
First, let's recall the chain rule for differentiation:
If h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x).
Given that f and g are functions and we have the values of f(x), g(x), f'(x), and g'(x) at x = 1 and x = 2, we can substitute these values into the chain rule formula to find h'(1).
At x = 1:
f(1) = 8
g(1) = 2
f'(1) = 1/3
g'(1) = -3
Plugging these values into the chain rule formula, we get:
h'(1) = f'(g(1)) * g'(1) = f'(2) * (-3)
Looking at the given values, we can see that f'(2) is not given directly. However, we can approximate it by using a linear approximation.
Using the given values at x = 1 and x = 2, we can find the slope of the line passing through these points, which will give us an approximate value of f'(2).
Slope = (f(2) - f(1)) / (2 - 1) = (3 - 8) / (2 - 1) = -5
Now we can use this approximate slope to find an approximation for f'(2):
f'(2) ≈ f'(1) + slope ≈ 1/3 - 5 = -14/3
Substituting this approximation and the given value of g'(1) into the chain rule formula, we get:
h'(1) = (-14/3) * (-3) = 14
So, the value of h'(1) is 14, not -6π as you initially mentioned.