Problem 69 - Calculus Volume 1

Question

The given limit represents the derivative of a function \(y=f(x)\) at \(x=a\). Find \(f(x)\) and \(a\). $$ \lim _{h \rightarrow 0} \frac{\left[3(2+h)^{2}+2\right]-14}{h} $$

Step-by-Step Solution

Answer

The function is \( f(x) = 3x^2 + 2 \) and \( a = 2 \).

Step 1: Recognize the Definition of Derivative

The given limit \( \lim _{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} \) represents the derivative of the function \( f(x) \) at \( x = a \). Here, it is given as \( \lim _{h \rightarrow 0} \frac{[3(2+h)^{2}+2]-14}{h} \).

Step 2: Observe the Structure of the Limit

The expression inside the limit resembles \( f(a+h) - f(a) \). Compare the expression \( 3(2+h)^2 + 2 \) with \( f(a+h) \), and the constant \( 14 \) with \( f(a) \).

Step 3: Equate to Identify \( f(x) \) and \( a \)

Recognize \( f(a) = 14 \) and \( f(a+h) = 3(2+h)^2 + 2 \). By expanding the polynomial, \( 3(2+h)^2 + 2 \) becomes \( 3(4 + 4h + h^2) + 2 = 12 + 12h + 3h^2 + 2 \). Set this equal to \( f(x) \) under the form \( f(x) = 3x^2 + 2 \).

Step 4: Determine the Value of \( a \)

Since \( f(a) = 14 \), this implies \( 3a^2 + 2 = 14 \). Solving for \( a \), we get \( 3a^2 = 12 \), thus \( a^2 = 4 \). Therefore, \( a = 2 \) or \( a = -2 \). However, the substitution \((2+h)\) in the original expression suggests \( a = 2 \).

Step 5: Confirm \( f(x) \) as a Correct Match

Substitute \( a = 2 \) in \( f(x) = 3x^2 + 2 \). When \( x = 2 \), \( f(2) = 3(2)^2 + 2 = 12 + 2 = 14 \), confirming our function and \( a \) are correct.

Key Concepts

Limit of a Function

In calculus, the concept of a limit is essential for understanding the behavior of functions as they approach specific points. The limit of a function, particularly as the input approaches zero, helps us understand how the function behaves as it gets infinitely close to a certain point.
When we say the limit of a function as the input approaches a particular value, we are interested in the value that the function tends to get closer to. In mathematical terms, for a function \(f(x)\), the limit as \(x\) approaches \(a\) is denoted by \(\lim_{x \to a} f(x)\).

• If the function gets infinitely close to a number, the limit exists at that number.
• If the function never settles near any number, the limit might not exist.

This concept is foundational for defining derivatives, which require understanding limits at approaching zero to determine the rate of change of functions.

Polynomial Functions

Polynomial functions are expressions consisting of variables and coefficients, structured as sums of terms of the form \(a_nx^n\), where \(n\) is a non-negative integer. These functions can include constants, linear terms, quadratic terms, and higher degree terms.
Polynomial functions have the general form \(f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\). They are simple yet powerful tools in mathematics because they can model a wide range of phenomena by varying the degree \(n\) and coefficients.

• When the degree is 2, we have a quadratic function, like in our exercise with \(3x^2 + 2\).
• Polynomials are continuous and smooth, making them ideal for calculus operations like differentiation.

In the given exercise, recognizing the polynomial structure helps us identify the function \(f(x)\) and understand how changes in \(x\) affect it.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function. This process is all about understanding how a function changes at any point, providing a way to calculate the slope of a function's graph at that point.
The derivative of a function \(f(x)\) is typically represented as \(f'(x)\) or \(\frac{df}{dx}\) and is found using the limit \[ \lim _{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} \]

• Differentiation tells us the rate of change of a quantity, such as speed or growth rate.
• In our exercise, differentiation is used to find how \(f(x)\) changes as we move away from \(x = a\).

Understanding how to differentiate polynomial functions is straightforward due to their simple power rule: for each term \(a_nx^n\), the derivative is \(n \cdot a_n x^{n-1}\). This makes polynomial differentiation highly systematic.

Derivative at a Point

The derivative at a specific point gives a precise rate of change of the function at that point. This rate is equivalent to the slope of the tangent line to the function's graph at the given point.
For the function \( f(x) \) at a point \( x = a \), the derivative \( f'(a) \) is thus \[ \lim _{h \rightarrow 0} \frac{f(a + h) - f(a)}{h} \]
This formula can be seen as observing what happens to \(f(x)\) as it shifts slightly from \(x = a\).

• The derivative at a point is crucial for applications such as determining the instantaneous velocity of an object at a moment in time.
• In the exercise, identifying \(a = 2\) helps find the effect small changes in \(x\) have on \(f(x)\) at \(x = 2\).

Understanding the derivative at a point helps solve real-world problems involving optimization and modeling dynamic systems.