Problem 72 - Calculus Volume 1

Question

The given limit represents the derivative of a function \(y=f(x)\) at \(x=a\). Find \(f(x)\) and \(a\). $$ \lim _{h \rightarrow 0} \frac{\left[2(3+h)^{2}-(3+h)\right]-15}{h} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the structure

The given expression is in the form of a limit that represents a derivative: \[\lim _{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}.\]Comparing this with the given limit, we can see that the expression that appears in place of \(f(x_0 + h)\) is \(2(3+h)^2 - (3+h)\), and \(f(x_0) = 15\).

Step 2: Expand the expression

Expand the expression \(2(3+h)^2 - (3+h)\):Start by expanding \((3+h)^2 = 9 + 6h + h^2\).Multiply by 2:\[2(3+h)^2 = 2(9 + 6h + h^2) = 18 + 12h + 2h^2.\]Subtract \((3+h)\):\[2(3+h)^2 - (3+h) = 18 + 12h + 2h^2 - 3 - h = 15 + 11h + 2h^2.\]

Step 3: Confirm the form with f(x)

Now recall that:\[f(x_0 + h) - f(x_0) = 15 + 11h + 2h^2 - 15 = 11h + 2h^2.\]So, the function \(f(x)\) as determined by the expression should evaluate to 15 when \(x=3\), which fits \(f(x) = 2x^2 - x\).

Step 4: Derivation of x-value a

Since \(f(x_0) = 15\), find \(x_0\) such that:\[2x_0^2 - x_0 = 15.\]Plug \(x_0 = 3\) into the equation:\[2(3)^2 - 3 = 18 - 3 = 15.\]Thus, \(x_0 = 3\) is consistent.

Key Concepts

Limit Calculation

The process of limit calculation is essential in understanding derivatives. A limit basically examines the behavior of a function as it approaches a specific point. In this context, it's like focusing the lens of a microscope on the function at a particular point.
So, what does the expression \(\lim _{h \rightarrow 0} \frac{[2(3+h)^{2}-(3+h)]-15}{h}\) mean? It represents the slope of the tangent to the curve at a particular point, which is the derivative. The limit signifies the change in \(y\) values over an infinitesimally small change in \(x\).

To solve this limit, follow these steps:

• Identify the structure of the limit, akin to recognizing a familiar pattern.
• Expand and simplify the expression within the limit to make things clearer.
• Evaluate the limit as \(h\) approaches zero, resolving the expression to find the derivative.

Understanding limit calculation enables you to transition smoothly to finding derivatives.

Differentiation

Differentiation is all about finding the derivative of a function. Think of it as discovering the rate at which things are changing. More formally, it's the process of determining the slope of the tangent line to a curve at any point.
In our exercise, we expanded \(2(3+h)^2 - (3+h)\) into \(18 + 12h + 2h^2 - 3 - h\) which further simplifies to \(15 + 11h + 2h^2\). This simplification prepares the expression for differentiation.

In practice, differentiation uses principles from limits to find how a function changes.

• First expansion happens: \((3+h)^2 = 9 + 6h + h^2\)
• Then multiplying by 2 gives: \(2(9 + 6h + h^2)\)
• Finally, subtract \((3+h)\).

Completion of this process allows for the final step of checking the derivative by comparing the form to general derivative knowledge. Differentiation turns calculus into a tool for exploring change.

Function Evaluation

Function evaluation seeks to find the value of the function at specific points, which is critical for verifying solutions. For this exercise, we needed to confirm \(f(x)\) and \(a\) such that \(f(a) = 15\).
Using the expression \(f(x) = 2x^2 - x\), we set \(x = 3\) for evaluation. By calculating this:

- Plug in \(x = 3\) into the function to get \(f(3) = 2(3)^2 - 3\)
- Simplify to find \(f(3) = 18 - 3 = 15\)

• This confirms that our \(x_0 = 3\) is true and accurate.
• This accurate calculation verified our evaluated expression fits the function properly.

Function evaluation checks that the points \(a\) and \(f(x)\) align, confirming the derivative setup suits the exercise's demand. Thus, it is vital for ensuring all derivative attempts comply with given conditions.