Chapter 3: Problem 59
Find the function The following limits represent the slope of a curve \(y=f(x)\) at the point \((a, f(a)) .\) Determine a possible function \(f\) and number \(a ;\) then calculate the limit. $$\lim _{h \rightarrow 0} \frac{(2+h)^{4}-16}{h}$$
Short Answer
Expert verified
Answer: The function is \(f(x) = (2+x)^4\) and the value of \(a\) is 0.
Step by step solution
01
1. Recognize the limit as the derivative of a function
The given limit has the form of the definition of the derivative of a function \(f(x)\). Recall that the definition of the derivative of a function \(f(x)\) at a point \(x=a\) is as follows:
$$f'(a) = \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$$
We are given:
$$\lim _{h \rightarrow 0} \frac{(2+h)^{4}-16}{h}$$
Now, we should compare the given limit to the definition of the derivative to find a matching function \(f(x)\) and value of \(a\).
02
2. Find a matching function and value of \(a\)
In our given limit expression, we can observe that \((2+h)^{4}\) has the form of \(f(a+h)\) and \(16\) has the form of \(f(a)\). So, we have:
$$f(a+h) = (2+h)^{4}$$
$$f(a) = 16$$
We can easily get the base function \(f(x)\) by replacing \(a+h\) with \(x\) in the expression \((2+h)^4\). This gives us:
$$f(x) = (2+x)^{4}$$
Now, we need to find the value of \(a\). Since \(f(a) = 16\), we can find \(a\) using our base function:
$$16 = (2+a)^{4}$$
Solving for \(a\), using the fourth root, gives us:
$$a = \sqrt[4]{16} - 2 = 2 - 2 = 0$$
03
3. Calculate the limit
Now that we have found the base function \(f(x) = (2+x)^4\) and the value of \(a=0\), we can calculate the limit using the given expression:
$$\lim _{h \rightarrow 0} \frac{(2+h)^{4}-16}{h} = \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} = f'(0)$$
We already know that \(f(x) = (2+x)^4\), so we can find \(f'(x)\) by taking the derivative using the power rule:
$$f'(x) = 4(2+x)^3$$
Now we can find \(f'(0)\):
$$f'(0) = 4(2+0)^3 = 4 \cdot 2^3 = 32$$
The given limit is the derivative of the function \(f(x) = (2+x)^4\) at the point \((0, f(0))\), and the limit evaluates to 32.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative of a Function
Understanding the derivative of a function is fundamental in calculus. It measures how a function's output value changes as the input changes. In simpler terms, think of it as a way to calculate the exact rate at which one quantity changes with respect to another.
For instance, if we have a function that represents the distance a car travels over time, the derivative of that function tells us the car's speed at any given moment. The derivative gives us the slope of the tangent line to the curve at any point, which is a precise measure of the rate of change at that specific point.
This concept is crucial because it allows us to understand motion and change in various physical and mathematical contexts, enabling us to predict and model real-world phenomena.
For instance, if we have a function that represents the distance a car travels over time, the derivative of that function tells us the car's speed at any given moment. The derivative gives us the slope of the tangent line to the curve at any point, which is a precise measure of the rate of change at that specific point.
This concept is crucial because it allows us to understand motion and change in various physical and mathematical contexts, enabling us to predict and model real-world phenomena.
Power Rule
The power rule is a basic technique used to find the derivative of a function that involves a power of x. The rule states that if you have a function of the form \( f(x) = x^n \), where n is a real number, the derivative of that function is \( f'(x) = nx^{n-1} \).
To apply the power rule to the exercise at hand, we look at the base function \( f(x) = (2+x)^4 \). According to the power rule, the derivative \( f'(x) \) is \( 4(2+x)^3 \). This simplifies the process of finding derivatives for polynomial functions significantly, making it one of the most widely used rules in differential calculus.
To apply the power rule to the exercise at hand, we look at the base function \( f(x) = (2+x)^4 \). According to the power rule, the derivative \( f'(x) \) is \( 4(2+x)^3 \). This simplifies the process of finding derivatives for polynomial functions significantly, making it one of the most widely used rules in differential calculus.
Slope of a Curve
The slope of a curve at any given point is the steepness or incline of the tangent line at that point. It tells us how sharp or gentle a curve is at a particular location. In mathematics, particularly in calculus, the slope of a curve is the same as the derivative at that point.
The concept of slope is directly connected to the incline of a hill or the steepness of a ramp. Mathematically, if we have a function \( y = f(x) \), the slope of the curve at a point \( x = a \) is represented by \( f'(a) \), which is the value of the derivative at that point. Having the ability to calculate the slope at any given point is essential in many fields, such as physics, economics, and engineering, because it helps predict and understand behavior in systems that change.
The concept of slope is directly connected to the incline of a hill or the steepness of a ramp. Mathematically, if we have a function \( y = f(x) \), the slope of the curve at a point \( x = a \) is represented by \( f'(a) \), which is the value of the derivative at that point. Having the ability to calculate the slope at any given point is essential in many fields, such as physics, economics, and engineering, because it helps predict and understand behavior in systems that change.
Definition of the Derivative
The formal definition of the derivative of a function at a point is the limit of the difference quotient as the interval approaches zero. Mathematically, it is expressed as \[ f'(a) = \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \].
This limit, if it exists, gives us the instantaneous rate of change of the function at the point \( a \). In the context of the given exercise, the limit \( \lim _{h \rightarrow 0} \frac{(2+h)^{4}-16}{h} \) represents the derivative of the function \( f(x) = (2+x)^4 \) at the point \( x = 0 \), which was calculated to be 32. Therefore, the derivative is essentially the slope of the tangent line to the function at a specific point and provides a powerful tool for analysis and problem-solving.
This limit, if it exists, gives us the instantaneous rate of change of the function at the point \( a \). In the context of the given exercise, the limit \( \lim _{h \rightarrow 0} \frac{(2+h)^{4}-16}{h} \) represents the derivative of the function \( f(x) = (2+x)^4 \) at the point \( x = 0 \), which was calculated to be 32. Therefore, the derivative is essentially the slope of the tangent line to the function at a specific point and provides a powerful tool for analysis and problem-solving.