Question

A function \(f\) and \(a\) value \(a\) are given. Approximate the limit of the difference quotient, \(\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h},\) using \(h=\pm 0.1, \pm 0.01 .\) $$ f(x)=-4 x^{2}+5 x-1, \quad a=-3 $$

Short answer

The approximate limit is 29.

Step-by-step solution

Identify the Function and the Point

The function given is \( f(x) = -4x^2 + 5x - 1 \) and the point at which we are finding the limit is \( a = -3 \).

Calculate the Function Values

Calculate \( f(a) = f(-3) \).\[ f(-3) = -4(-3)^2 + 5(-3) - 1 = -4(9) - 15 - 1 = -36 - 15 - 1 = -52 \]

Express the Difference Quotient

The difference quotient is given by:\[ \frac{f(a+h) - f(a)}{h} \]We need to calculate \( f(a+h) = f(-3+h) \).

Compute f(a+h) for Positive h Values

First compute \( f(-3 + h) \) for \( h = 0.1 \) and \( h = 0.01 \).- For \( h=0.1 \): \[ f(-3 + 0.1) = f(-2.9) = -4(-2.9)^2 + 5(-2.9) - 1 \] \[ = -4 imes 8.41 + 5(-2.9) - 1 = -33.64 - 14.5 - 1 = -49.14 \]- For \( h=0.01 \): \[ f(-3 + 0.01) = f(-2.99) = -4(-2.99)^2 + 5(-2.99) - 1 \] \[ = -4 imes 8.9401 + 5(-2.99) - 1 = -35.7604 - 14.95 - 1 = -51.7104 \]

Compute the Difference Quotient for Positive h

Compute the difference quotient for positive values of \( h \), using the formula obtained earlier:- For \( h=0.1 \): \[ \text{Difference quotient} = \frac{-49.14 - (-52)}{0.1} = \frac{2.86}{0.1} = 28.6 \]- For \( h=0.01 \): \[ \text{Difference quotient} = \frac{-51.7104 - (-52)}{0.01} = \frac{0.2896}{0.01} = 28.96 \]

Compute f(a+h) for Negative h Values

Now calculate \( f(a+h) \) for \( h = -0.1 \) and \( h = -0.01 \).- For \( h=-0.1 \): \[ f(-3 -0.1) = f(-3.1) = -4(-3.1)^2 + 5(-3.1) - 1 \] \[ = -4 imes 9.61 + 5(-3.1) - 1 = -38.44 - 15.5 - 1 = -54.94 \]- For \( h=-0.01 \): \[ f(-3 -0.01) = f(-3.01) = -4(-3.01)^2 + 5(-3.01) - 1 \] \[ = -4 imes 9.0601 + 5(-3.01) - 1 = -36.2404 - 15.05 - 1 = -52.2904 \]

Compute the Difference Quotient for Negative h

Compute the difference quotient for negative values of \( h \).- For \( h=-0.1 \): \[ \text{Difference quotient} = \frac{-54.94 - (-52)}{-0.1} = \frac{-2.94}{-0.1} = 29.4 \]- For \( h=-0.01 \): \[ \text{Difference quotient} = \frac{-52.2904 - (-52)}{-0.01} = \frac{-0.2904}{-0.01} = 29.04 \]

Approximate the Limit

From the calculated difference quotients, we see:- For \( h=0.1 \) and \( h=-0.1 \), the values are 28.6 and 29.4 respectively.- For \( h=0.01 \) and \( h=-0.01 \), the values are 28.96 and 29.04 respectively.The approximate limit of the difference quotient as \( h \rightarrow 0 \) is 29.

Key concepts

Difference Quotient

The difference quotient is a crucial concept in calculus, often used to find the instantaneous rate of change of a function, which leads to the definition of a derivative. It is expressed by the formula:

• \[ \frac{f(a+h) - f(a)}{h} \]

This formula represents the average rate of change of the function \(f(x)\) over the interval from \(a\) to \(a + h\). As \(h\) becomes very small, the difference quotient gives an approximation of the derivative of \(f(x)\) at \(a\).

In practice, as shown in the exercise, we vary \(h\) to compute the difference quotient for both positive and negative values. This offers insight into how the function behaves around the point \(a\). When \(h\) approaches zero, this quotient tends toward the function's derivative at that point. Hence, understanding this concept is foundational for grasping the behavior and changes in functions.

Limit Approximation

Limit approximation is an essential technique in calculus used to estimate the value of a function as its variables approach a certain point, typically zero. It's vital when calculating derivatives using the difference quotient.

In the exercise, you approximate the limit of the difference quotient as \(h\) approaches zero. This involves evaluating the function at small, incremental values of \(h\), such as \(\pm 0.1\) and \(\pm 0.01\). By examining how the difference quotient changes for these small values of \(h\), you can deduce the behavior of the function's derivative at a given point.

• Positive \(h\) values provide insight from one direction.
• Negative \(h\) values offer perspective from the opposite direction.

As you decrease \(h\), the approximation becomes more accurate. The exercise concludes that the limit of this difference as \(h\) draws near zero is an approximation of the derivative, which helps in understanding the rate of change of the function around the point \(a\).

Polynomial Function

Polynomial functions are a fundamental component of algebra and calculus, characterized by their structure of terms consisting of constants, variables, and non-negative integer exponents. The given function in the exercise is:

• \[ f(x) = -4x^2 + 5x - 1 \]

This is a quadratic polynomial, notable for its parabolic graph and predictable behavior. Each term impacts the function's curvature and direction. Understanding this function's nature helps in performing accurate calculations.

In particular, such functions are advantageous because their derivatives can be calculated consistently using well-known rules. The exercise involves substituting different values into this polynomial to estimate the derivative by computing the difference quotient.

Polynomial function behavior is relatively straightforward, making it an excellent starting point for learning calculus concepts such as limits and derivatives.

Derivative Estimation

Derivative estimation uses the concepts of limit and difference quotient to find the slope of a function at a particular point. This slope is fundamental in calculus, representing the instantaneous rate of change.

In the exercise, the approximation process begins by computing the difference quotient with various \(h\) values, both positive and negative. As \(h\) approaches zero - a fundamental technique known as limit approximation - the values of the difference quotient converge towards the derivative of the function at \(a\).

Estimation is key when exact calculations are complex or when dealing with real-world data. By understanding how the function shifts as \(h\) changes, you can effectively find the tangent's slope to the curve at point \(a\).

• This process uncovers the gradient of the function at specific points.
• It equips you with the ability to predict and comprehend dynamic shifts within the function.

The derivative thus serves as a powerful tool for analysis, allowing us to deeply understand the behavior and trends of polynomial functions, such as the one given in the exercise.