Problem 66 - Brief Applied Calculus

Question

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ f(x)=3 x^{2}-5 x+2 $$

Step-by-Step Solution

Answer

The simplified expression is \( 6x + 3h - 5 \).

Step 1: Understand the Function

We are given a quadratic function \( f(x) = 3x^2 - 5x + 2 \). We need to find the expression \( \frac{f(x+h)-f(x)}{h} \). This expression is crucial for understanding the rate of change of the function and is related to the derivative.

Step 2: Substitute \( x+h \) into the Function

Substitute \( x+h \) into the function \( f(x) \):\[ f(x+h) = 3(x+h)^2 - 5(x+h) + 2 \]Simplify this expression by expanding the binomials.

Step 3: Expand and Simplify \( f(x+h) \)

First expand \((x+h)^2\) to get \(x^2 + 2xh + h^2\). Then substitute:\[ f(x+h) = 3(x^2 + 2xh + h^2) - 5(x+h) + 2 \]Simplify further:\[ = 3x^2 + 6xh + 3h^2 - 5x - 5h + 2 \].

Step 4: Find \( f(x+h) - f(x) \)

Calculate \( f(x+h) - f(x) \) by substituting in the expressions for \( f(x+h) \) and \( f(x) \):\[ f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 5x - 5h + 2) - (3x^2 - 5x + 2) \]Simplify:\[ = 6xh + 3h^2 - 5h \].

Step 5: Divide by \( h \) and Simplify

Now divide the expression from Step 4 by \( h \):\[ \frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 5h}{h} \]Factor \( h \) out from the numerator:\[ = \frac{h(6x + 3h - 5)}{h} \]Cancel out \( h \) to simplify:\[ = 6x + 3h - 5 \].

Step 6: Conclusion

The expression \( \frac{f(x+h)-f(x)}{h} \) simplifies to \( 6x + 3h - 5 \). This shows the average rate of change of the function around \( x \) as \( h \) approaches 0.

Key Concepts

Quadratic Function

A quadratic function is a polynomial function of degree two, generally expressed in the form \( f(x) = ax^2 + bx + c \). It is characterized by the presence of the term \( x^2 \), which imparts a parabolic shape to the graph of the function. In our specific problem, the quadratic function given is \( f(x) = 3x^2 - 5x + 2 \).Key features of quadratic functions include:

• The vertex, which is the highest or lowest point of the parabola depending on the sign of \( a \).
• The axis of symmetry, a vertical line that passes through the vertex and divides the parabola into two mirror images.
• The direction of the parabola, which opens upwards if \( a > 0 \) and downwards if \( a < 0 \).

Quadratic functions appear frequently in various fields like physics, engineering, and economics because they model relationships where variables change under constraints, describing paths of projectiles, profit optimization, and more.

Rate of Change

The rate of change is a fundamental concept that describes how a function's output changes with respect to changes in the input. For a function \( f(x) \), the rate of change over an interval is determined by the difference quotient, which is expressed as \( \frac{f(x+h) - f(x)}{h} \).In the context of the given quadratic function, the simplified difference quotient \( 6x + 3h - 5 \) is used. This expression tells us how the output of the quadratic changes as \( h \) changes, providing an average rate of change over a small interval around \( x \).Understanding the rate of change:

• Helps us anticipate the behavior of the function over an interval.
• Indicates how fast or slow the function's value is increasing or decreasing.

The rate of change becomes especially valuable when examining trends and patterns or making predictions based on the function's graph.

Derivative

The derivative of a function is a central concept in calculus that represents the instantaneous rate of change of the function with respect to one of its variables. When \( h \) approaches zero, the difference quotient \( \frac{f(x+h) - f(x)}{h} \) transforms into the derivative.In relation to our quadratic function, as \( h \to 0 \), the expression \( 6x + 3h - 5 \) simplifies to \( 6x - 5 \). This is the derivative of the quadratic function \( f(x) = 3x^2 - 5x + 2 \).Key insights from derivatives:

• They provide critical information about the function's increasing or decreasing behavior at a specific point.
• Derivatives allow us to find the slope of the tangent to the function's graph at any point \( x \).
• By identifying where the derivative equals zero, we can locate the function's critical points, which help determine maxima, minima, or inflection points.

Mastering the derivative concept is essential for tackling more complex problems in calculus and making informed decisions in applied mathematics settings.