Problem 22
Question
For Problems 21 through 24 use the limit definition of \(f^{\prime}(a)\) to find the derivative of \(f\) at the point indicated. $$ f(x)=x(x+3) \text { at } x=2 $$
Step-by-Step Solution
Verified Answer
The derivative of the function \(f(x)=x(x+3)\) at the point \(x=2\) is 5.
1Step 1: Write down the formula for derivative
The limit definition of the derivative of a function f at a specific point \(a\) can be written as:\[f^{\prime}(a) = \lim_{{h} \to 0} \frac {f(a + h) - f(a)}{h}\]Now we know that \(f(x)=x(x+3)\) and the point at which we need to find the derivative is \(x=2\)
2Step 2: Substitute into the formula
To find the derivative, we substitute our function, \(f(x)\), and the value of \(x=2\) into our derivative formula. \[f^{\prime}(2) = \lim_{{h} \to 0} \frac {(2+h)((2+h) + 3) - 2(2+3)}{h}\]
3Step 3: Simplify the equation
Simplify the equation by solving the expression inside the brackets.\[f^{\prime}(2) = \lim_{{h} \to 0} \frac {5h + h^{2}}{h}\]Now, divide each of the terms in the numerator by \(h\)\[f^{\prime}(2) = \lim_{{h} \to 0} (5 + h)\]
4Step 4: Calculate the limit
As \(h\) tends to 0, our function becomes \[f^{\prime}(2) = 5 + 0 = 5\]
Key Concepts
Derivative CalculationPolynomial FunctionsLimits in Calculus
Derivative Calculation
Derivatives are fundamental in calculus and are used to determine how a function changes at a given point. The derivative essentially measures the "rate of change" of a function's output relative to its input. To calculate a derivative using the limit definition, we use the formula: \[ f^{\prime}(a) = \lim_{{h} \to 0} \frac {f(a + h) - f(a)}{h} \] The core idea is to find the instantaneous rate of change by examining how the function behaves over an infinitesimally small interval. This contrasts with average rates of change, which consider larger intervals. For the given problem, we apply this to the function \( f(x) = x(x+3) \) at \( x = 2 \). By substituting and simplifying, we aim to express the rate of change in the simplest form as \( h \) approaches zero, ultimately leading to the derivative \( f^{\prime}(2) = 5 \).
- Derivatives tell us how fast a function changes at any point.
- The limit definition involves expressions shrinking to near zero.
- Being comfortable with simplification and algebraic manipulation is key.
Polynomial Functions
Polynomial functions are expressions consisting of variables and coefficients, involving only addition, subtraction, multiplication, and non-negative integer exponents. The function given, \( f(x) = x(x+3) \), is a quadratic polynomial:
- It can be expanded to \( f(x) = x^2 + 3x \).
Polynomial functions are smooth and continuous, which makes them ideal for differentiation, as their derivatives are not complex to find. Each term independently contributes to the final derivative, allowing for straightforward calculations:
- The power rule aids in finding derivatives of individual terms, making polynomials particularly simple to differentiate term by term.
- It can be expanded to \( f(x) = x^2 + 3x \).
Polynomial functions are smooth and continuous, which makes them ideal for differentiation, as their derivatives are not complex to find. Each term independently contributes to the final derivative, allowing for straightforward calculations:
- The power rule aids in finding derivatives of individual terms, making polynomials particularly simple to differentiate term by term.
- Quadratic polynomials involve terms up to \( x^2 \).
- They're continuous and differentiable across all real numbers.
- Power rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \), a foundational tool in calculus.
Limits in Calculus
Limits are a fundamental concept in calculus and provide a framework to analyze behavior as inputs approach a specific point. Using limits, calculus can handle notions of continuity, derivatives, and integrals. In the context of derivative calculation using limits, like in our problem, the pivotal aspect is:
- Evaluating \( \lim_{{h} \to 0} \) maximizes precision, modeling infinitely small changes in \( x \).
A core part of obtaining the derivative \( f'(x) \) is to identify the behavior of \( \frac {f(a+h) - f(a)}{h} \) as \( h \) approaches zero. Limits help conceptualize change at a micro level, bridging the gap between algebra and calculus for tangible real-world applications, like motion or growth rates.
- Evaluating \( \lim_{{h} \to 0} \) maximizes precision, modeling infinitely small changes in \( x \).
A core part of obtaining the derivative \( f'(x) \) is to identify the behavior of \( \frac {f(a+h) - f(a)}{h} \) as \( h \) approaches zero. Limits help conceptualize change at a micro level, bridging the gap between algebra and calculus for tangible real-world applications, like motion or growth rates.
- Limits underscore continuous change analysis.
- They enable the transition from difference quotients to precise derivatives.
- Conceptually, they refine our approach to understanding dynamic systems.
Other exercises in this chapter
Problem 19
Let \(f(x)=x^{-\frac{1}{2}}\). Use the limit de nition of derivative to show that \(f^{\prime}(x)=-\frac{1}{2} x^{-\frac{3}{2}}\).
View solution Problem 21
For Problems 21 through 24 use the limit definition of \(f^{\prime}(a)\) to find the derivative of \(f\) at the point indicated. $$ f(x)=\frac{3}{2-x} \text { a
View solution Problem 23
For Problems 21 through 24 use the limit definition of \(f^{\prime}(a)\) to find the derivative of \(f\) at the point indicated. $$ f(x)=\frac{x}{2}+\frac{2}{x}
View solution Problem 24
For Problems 21 through 24 use the limit definition of \(f^{\prime}(a)\) to find the derivative of \(f\) at the point indicated. $$ f(x)=(x-3)^{2} \text { at }
View solution