/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A table of values for \(f, g, f^... [FREE SOLUTION] | 91Ó°ÊÓ

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A table of values for \(f, g, f^{\prime},\) and \(g^{\prime}\) is given. \(\begin{array}{|c|c|c|c|c|}\hline x & {f(x)} & {g(x)} & {f^{\prime}(x)} & {g^{\prime}(x)} \\ \hline 1 & {3} & {2} & {4} & {6} \\ {2} & {1} & {8} & {5} & {7} \\ {3} & {7} & {2} & {7} &{9}\\\\\hline\end{array}\) (a)If $$h(x)=f(g(x)), find h^{\prime}(1)$$ (b)If $$H(x)=g(f(x)), find H^{\prime}(1)$$.

Short Answer

Expert verified
(a) \( h'(1) = 30 \); (b) \( H'(1) = 36 \).

Step by step solution

01

Understanding the Function Composition

To find the derivative of a composite function such as \( h(x) = f(g(x)) \), we use the chain rule. The chain rule states that the derivative of \( h(x) \) is \( h'(x) = f'(g(x)) \cdot g'(x) \). We will apply this formula to find \( h'(1) \).
02

Evaluating the Inner Function at \( x = 1 \)

For \( h(x) = f(g(x)) \), first evaluate \( g(1) \). From the table, we have \( g(1) = 2 \).
03

Applying the Chain Rule for \( h'(1) \)

Plug \( g(1) = 2 \) into the chain rule formula. We need \( f'(g(1)) \) and \( g'(1) \). From the table, \( f'(2) = 5 \) and \( g'(1) = 6 \). Therefore, \( h'(1) = f'(2) \cdot g'(1) = 5 \cdot 6 = 30 \).
04

Understanding the Function Composition for H

For \( H(x) = g(f(x)) \), we again use the chain rule: \( H'(x) = g'(f(x)) \cdot f'(x) \). We will apply this formula to find \( H'(1) \).
05

Evaluating the Inner Function for \( H \) at \( x = 1 \)

For \( H(x) = g(f(x)) \), first evaluate \( f(1) \). From the table, \( f(1) = 3 \).
06

Applying the Chain Rule for \( H'(1) \)

Substitute \( f(1) = 3 \) into the chain rule formula. We need \( g'(f(1)) \) and \( f'(1) \). From the table, \( g'(3) = 9 \) and \( f'(1) = 4 \). Therefore, \( H'(1) = g'(3) \cdot f'(1) = 9 \cdot 4 = 36 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composite Functions
Composite functions are like Russian dolls. You have one function nestled inside another. When we say a function like this is composite, it means the output of one function becomes the input for another. In the case of the exercise, we deal with two types of composite functions: \( h(x) = f(g(x)) \) and \( H(x) = g(f(x)) \).

Think of \( g(x) \) as our inner function - it processes the input first. Then, its result goes into \( f \) or vice versa, making it a nested evaluation. Composite functions like these are common in mathematics and are used to model real-world situations where multiple stages of change or processes occur.

To evaluate or differentiate such functions, always keep an eye on the layers. Remember: tackle the inner function first before the outer one. It's like peeling an onion - one layer at a time.
Derivative Calculations
In the realm of calculus, derivative calculations are vital for understanding change. They tell us how a function's output changes as its input changes. When dealing with composite functions, like our \( h(x) \) and \( H(x) \), these calculations require special attention using a method known as the chain rule.

The chain rule is your go-to tool when you need to find derivatives of composite functions. It connects the rates of change of the inner and outer functions. For \( h(x) = f(g(x)) \), the derivative, using the chain rule, is \( h'(x) = f'(g(x)) \cdot g'(x) \). Similarly, for \( H(x) = g(f(x)) \), it's \( H'(x) = g'(f(x)) \cdot f'(x) \).

Here’s a handy tip:
  • Always evaluate the inner function first and find its rate of change or derivative.
  • Then, apply the chain rule: multiply the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function.
This systematic approach makes sure you get the right results as we saw with finding \( h'(1) \) and \( H'(1) \).

Whether you’re solving real-world problems or theoretical exercises, understanding derivatives gives you the power to analyze how changes ripple through composite systems.
Function Composition
Function composition is all about understanding how functions interact when one is applied to the result of another. It’s like cooking - each function represents a step or ingredient that transforms your initial input into something new.

When composing two functions, you write it as \( (f \circ g)(x) \), meaning \( f(g(x)) \). It’s not just about operating on access; it’s about sequencing transformations, where the output of \( g(x) \) is fed into \( f(x) \).

To master function composition, consider:
  • The order of operations matters: evaluate the inner function first for proper sequencing.
  • Understand that composition can change the domain and range of a function, as the allowable inputs to the final function depend on the result of the inner function.
In our exercise, this is crucial for determining points such as \( g(1) \) or \( f(1) \) before proceeding to further calculations.

Practicing with different functions will solidify your grasp of how to manage function composition effectively, making this concept a powerful tool in both pure and applied mathematical contexts.

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Most popular questions from this chapter

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