Problem 3
Question
Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a calculator. \(\sqrt[3]{124}\)
Step-by-Step Solution
Verified Answer
The approximation of \( \sqrt[3]{124} \) using linear approximation is 4.9867, close to the calculator value 4.989.
1Step 1: Identify the function and the point 'a'
The function we are dealing with is the cube root function, so we define it as \( f(x) = x^{1/3} \). To use linear approximation, select a value of 'a' close to 124 for which the cube root is easy to compute. We choose \( a = 125 \) since \( \sqrt[3]{125} = 5 \).
2Step 2: Calculate the derivative of the function
Find the derivative of \( f(x) = x^{1/3} \). By using the power rule for derivatives, the derivative \( f'(x) \) is \( \frac{1}{3}x^{-2/3} \). Simplifying, we have \( f'(x) = \frac{1}{3x^{2/3}} \).
3Step 3: Evaluate the function and its derivative at 'a'
Calculate \( f(a) \) at \( a = 125 \). We have \( f(125) = \sqrt[3]{125} = 5 \). Also calculate the derivative at \( a = 125 \): \( f'(125) = \frac{1}{3 \times 125^{2/3}} = \frac{1}{75} \), since \( 125^{2/3} = 25 \).
4Step 4: Write and evaluate the linear approximation formula
Use the linear approximation formula: \[ f(x) \approx f(a) + f'(a)(x-a) \] Substitute the values: \[ \sqrt[3]{124} \approx 5 + \frac{1}{75}(124 - 125) \] Simplify to get:\[ 5 + \frac{1}{75}(-1) = 5 - \frac{1}{75} \approx 5 - 0.0133 = 4.9867 \]
5Step 5: Compare with calculator result
Using a calculator, find \( \sqrt[3]{124} \approx 4.989 \). The linear approximation we found was 4.9867, which is very close to the calculator value.
Key Concepts
Derivative CalculationPower RuleCube Root Function
Derivative Calculation
In calculus, finding the derivative of a function is a fundamental concept. The derivative represents the rate of change of a function with respect to a variable. To calculate the derivative, different rules can be applied based on the form of the function.
In this exercise, we dealt with the cube root function expressed as a power: \( f(x) = x^{1/3} \). Calculating the derivative of this function involves using the power rule. Understanding how to find derivatives is crucial for applications like curve sketching and determining instantaneous rates of change.
In this exercise, we dealt with the cube root function expressed as a power: \( f(x) = x^{1/3} \). Calculating the derivative of this function involves using the power rule. Understanding how to find derivatives is crucial for applications like curve sketching and determining instantaneous rates of change.
Power Rule
The power rule is one of the simplest and most widely used techniques for finding the derivative of functions that are powers of \( x \). The rule states that for any function \( f(x) = x^n \), the derivative \( f'(x) \) is \( nx^{n-1} \).
Understanding this rule helps in quickly computing derivatives for polynomial functions and even more complex expressions, such as roots and fractional powers. Here, for the cube root function \( x^{1/3} \), the derivative calculation follows the power rule, resulting in:\[ f'(x) = \frac{1}{3}x^{-2/3} \]
The power rule simplifies many calculus problems by providing a straightforward method to find the rate of change instantly.
Understanding this rule helps in quickly computing derivatives for polynomial functions and even more complex expressions, such as roots and fractional powers. Here, for the cube root function \( x^{1/3} \), the derivative calculation follows the power rule, resulting in:\[ f'(x) = \frac{1}{3}x^{-2/3} \]
The power rule simplifies many calculus problems by providing a straightforward method to find the rate of change instantly.
Cube Root Function
The function \( f(x) = \sqrt[3]{x} \) is a fundamental example of a root function, specifically the cube root function. This type of function is defined for all real numbers and provides the value that, when raised to the third power, produces \( x \).
Cube roots are useful in many areas of mathematics, including solving cubic equations and analyzing growth models. In our exercise, we approximated \( \sqrt[3]{124} \) using linear approximation. This method is particularly beneficial when dealing with numbers that are difficult to compute directly, offering a simple way to estimate function values near a chosen point \( a \).
Overall, cube root functions, along with methods like linear approximations, expand our mathematical toolkit, enabling the solving of more complex problems beyond basic arithmetic.
Cube roots are useful in many areas of mathematics, including solving cubic equations and analyzing growth models. In our exercise, we approximated \( \sqrt[3]{124} \) using linear approximation. This method is particularly beneficial when dealing with numbers that are difficult to compute directly, offering a simple way to estimate function values near a chosen point \( a \).
Overall, cube root functions, along with methods like linear approximations, expand our mathematical toolkit, enabling the solving of more complex problems beyond basic arithmetic.
Other exercises in this chapter
Problem 2
In Problems \(1-8\), find \(\frac{d y}{d x}\) by implicit differentiation. $$ y=x^{2}+3 y x $$
View solution Problem 2
In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(x)=(4 x+5)^{3} $$
View solution Problem 3
Find the derivative at the indicated point from the graph of \(y=f(x)\). \(f(x)=2 x-3 ; x=-1\)
View solution Problem 3
Find the derivative with respect to the independent variable. $$ f(x)=3 \sin x+5 \cos x $$
View solution