Problem 63 - Brief Applied Calculus

Question

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ f(x)=5 x^{2} $$

Step-by-Step Solution

Answer

The simplified form is \(10x + 5h\).

Step 1: Express f(x+h)

Plug in \(x+h\) into the function. Since \(f(x) = 5x^2\), we find \(f(x+h)\) by replacing \(x\) with \(x+h\):\[f(x+h) = 5(x+h)^2\].

Step 2: Simplify f(x+h)

Expand \(5(x+h)^2\) using the formula \((a+b)^2 = a^2 + 2ab + b^2\):\[5(x+h)^2 = 5(x^2 + 2xh + h^2) = 5x^2 + 10xh + 5h^2\].

Step 3: Setup the Difference Quotient

Substitute \(f(x+h)\) and \(f(x)\) into the difference quotient:\[\frac{f(x+h) - f(x)}{h} = \frac{5x^2 + 10xh + 5h^2 - 5x^2}{h}\].

Step 4: Simplify the Difference Quotient

Cancel out \(5x^2\) and simplify the numerator:\[\frac{10xh + 5h^2}{h}\].

Step 5: Factor and Reduce

Factor out \(h\) from the numerator:\[\frac{h(10x + 5h)}{h}\]. Now, cancel \(h\) in the numerator and the denominator:\[10x + 5h\].

Step 6: Simplified Expression

The simplified form of the difference quotient is:\[10x + 5h\].

Key Concepts

Difference Quotient

The difference quotient is a fundamental tool in calculus and is used to define the derivative of a function. It measures the rate of change of the function as the input changes by a small amount, in this case represented by \(h\). To find the difference quotient for a function \(f(x)\), you calculate \(\frac{f(x+h) - f(x)}{h}\).

• Start by finding \(f(x+h)\), which is the value of the function at \(x+h\).
• Subtract \(f(x)\) from \(f(x+h)\) to find the change in the function.
• Divide by \(h\) to find how much the function changes per unit of \(h\).

The difference quotient simplifies to a form that gives insight into the rate of change, which is closely linked to the derivative.

Functions

A function in mathematics is a relation between a set of inputs and a set of permissible outputs, with each input being related to exactly one output. In this exercise, the function is \(f(x) = 5x^2\). This is a quadratic function, where:

• \(5\) is the coefficient that scales the input squared.
• The exponent \(2\) indicates it is a quadratic function.
• \(x\) is the variable of the function, its input.

When working with functions, you often need to evaluate them at different points, such as \(x\), \(x+h\), and others. Understanding how changes in the input affect the output is essential for comprehending calculus concepts.

Simplification

Simplification in mathematical expressions involves making the expression easier to interpret and understand by reducing it to its most basic form. During simplification, algebraic manipulations such as expanding, factoring, and cancelling terms are performed:

• **Expanding**: Use algebraic identities like \((a+b)^2 = a^2 + 2ab + b^2\) to expand expressions.
• **Factoring**: Find common factors to simplify components of the expression.
• **Cancelling Terms**: Remove common terms in fractions to reduce the expression.

In our example, we expanded \(5(x+h)^2\) to \(5x^2 + 10xh + 5h^2\), then cancelled and factored to ultimately reduce the difference quotient to a simplified form.

Derivatives

Derivatives are a core concept in calculus and involve finding the instantaneous rate of change of a function, commonly known as the slope of a function at a point. The derivative gives us insights into the behavior of functions and helps in understanding how changes at one variable affect another.

• The derivative at a point is the limit of the difference quotient as \(h\) approaches zero.
• In problems like these, the final simplified form of the difference quotient, when \(h\) is negligible, leads to the derivative expression \(10x\) for the function \(f(x) = 5x^2\).
• This means that the rate of change of the function \(f(x)\) at any point \(x\) is \(10x\).

Understanding derivatives is crucial because it forms the foundation for many applications of calculus, such as in physics, engineering, and economics.