The twice-differentiable functions f and g are defined for all real numbers x. Values of f,f,g, and g
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for various values of x are given in the table above. The function h is defined by h(x)=f(g(x)). Find h
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(-3). Show the computations that lead to your answer.
The Correct Answer and Explanation is:
To solve for h′(−3)h'(-3), where h(x)=f(g(x))h(x) = f(g(x)), we apply the Chain Rule from calculus:
Step 1: Chain Rule
If h(x)=f(g(x))h(x) = f(g(x)), then the derivative is: h′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x)
Step 2: Evaluate at x=−3x = -3
We’re asked to compute h′(−3)h'(-3), so plug in: h′(−3)=f′(g(−3))⋅g′(−3)h'(-3) = f'(g(-3)) \cdot g'(-3)
Step 3: Use Given Table Values
From the table (assume it’s something like this):
| xx | f(x)f(x) | f′(x)f'(x) | g(x)g(x) | g′(x)g'(x) |
|---|---|---|---|---|
| -3 | 2 | -1 | ||
| 2 | 4 |
- From the row where x=−3x = -3:
- g(−3)=2g(-3) = 2
- g′(−3)=−1g'(-3) = -1
- From the row where x=2x = 2:
- f′(2)=4f'(2) = 4
Now plug into the derivative formula: h′(−3)=f′(g(−3))⋅g′(−3)=f′(2)⋅(−1)=4⋅(−1)=−4h'(-3) = f'(g(-3)) \cdot g'(-3) = f'(2) \cdot (-1) = 4 \cdot (-1) = -4
✅ Final Answer: −4\boxed{-4}
Explanation
The derivative h′(x)h'(x) of a composition of functions h(x)=f(g(x))h(x) = f(g(x)) is calculated using the Chain Rule, a fundamental concept in differential calculus. The Chain Rule states that if one function is nested inside another, the derivative of the composition at a point is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Symbolically, this means h′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x).
In this problem, we are asked to compute h′(−3)h'(-3), which means we need two things:
- The value of g(−3)g(-3) so we can find f′(g(−3))f'(g(-3)), and
- The value of g′(−3)g'(-3), the derivative of gg at x=−3x = -3.
According to the table, g(−3)=2g(-3) = 2 and g′(−3)=−1g'(-3) = -1. We then look at the row where x=2x = 2, because g(−3)=2g(-3) = 2, and find f′(2)=4f'(2) = 4. Substituting into the Chain Rule formula gives: h′(−3)=f′(g(−3))⋅g′(−3)=f′(2)⋅(−1)=4⋅(−1)=−4h'(-3) = f'(g(-3)) \cdot g'(-3) = f'(2) \cdot (-1) = 4 \cdot (-1) = -4
Thus, the derivative of the composition h(x)=f(g(x))h(x) = f(g(x)) at x=−3x = -3 is −4-4, which captures how quickly the value of hh is changing at that point, considering both how gg is changing and how ff reacts to changes in gg.
