91Ó°ÊÓ

The concept of a derivative is fundamental in calculus, serving as a tool to measure how a function changes. To deeply understand this, we first explore the limit definition of a derivative. This definition answers the question, "How steep is the graph of the function at a point?".

The derivative of a function \( f \) at any point \( x \) is expressed through limits as:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

This formula represents the derivative as the slope of the tangent line to the curve at a particular point.
The expression \( f(x+h) - f(x) \) helps us understand how much the function's value changes over a tiny interval \( h \), and dividing by \( h \) provides the rate of change. Hence, the derivative is about understanding instantaneous rates of change.

It's like checking the speed of a car at an exact moment on its journey, not just the average speed over time.

A polynomial function is a mathematical expression consisting of variables, coefficients, and the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. In our problem, we are dealing with a specific polynomial function, \( f(x) = 4x^2 + 1 \). This function is a quadratic polynomial, the simplest form of a polynomial function involving squares.

Polynomial functions are crucial in calculus because they are smooth and continuous, which means derivatives can be calculated at any point without interruption.

• Coefficients like \( 4 \) in \( 4x^2 \) determine the stretch or compression of the graph.
• The constant term \( +1 \) shifts the graph vertically.

Understanding polynomial functions helps when applying the limit definition of the derivative because it becomes easier to apply algebraic simplifications. The example function, \( 4x^2 + 1 \), is typical of simple calculus problems aimed at teaching fundamental operations.

Once we have the derivative function using the limit definition, the next step is to evaluate it at specific points. Evaluating derivatives at a point helps us understand the behavior of the function at precise locations.

For our example function, the derivative obtained is \( f'(x) = 8x \). To find the derivative at specific points, we'll substitute the desired \( a \) value into the derivative function \( f'(x) \). Here’s how it works:

• For \( a = 2 \), substitute \( 2 \) into \( f'(x) \), giving us \( f'(2) = 8 \times 2 = 16 \).
• For \( a = 4 \), substitute \( 4 \), resulting in \( f'(4) = 8 \times 4 = 32 \).

By performing these substitutions, we retrieve the slope of the tangent to the curve at those specific points. This practical application of derivatives to evaluate the steepness where \( x = 2 \) and \( x = 4 \) provides a clearer view of the function's behavior.