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 \(h=\pm 0.1, \pm 0.01 .\) $$ f(x)=x^{2}+3 x-7, \quad a=1 $$

Short answer

The limit of the difference quotient is approximately 5.0.

Step-by-step solution

Calculate for h = 0.1

Plug in \(f(x) = x^2 + 3x - 7\) and \(a = 1\) into the difference quotient formula where \(h = 0.1\). Substitute: \(f(1+0.1) = (1.1)^2 + 3(1.1) - 7\). Calculate: \((1.1)^2 = 1.21\), \(3 \times 1.1 = 3.3\), \(f(1.1) = 1.21 + 3.3 - 7 = -2.49\). Now compute the difference quotient: \(\frac{f(1.1) - f(1)}{0.1} = \frac{-2.49 - (-3)}{0.1} = 5.1\).

Calculate for h = -0.1

Now use \(h = -0.1\) in the difference quotient formula.Substitute: \(f(1-0.1) = (0.9)^2 + 3(0.9) - 7\). Calculate: \((0.9)^2 = 0.81\), \(3 \times 0.9 = 2.7\), \(f(0.9) = 0.81 + 2.7 - 7 = -3.49\). Compute the difference quotient: \(\frac{f(0.9) - f(1)}{-0.1} = \frac{-3.49 - (-3)}{-0.1} = 4.9\).

Calculate for h = 0.01

Use \(h = 0.01\) in the difference quotient formula.Substitute: \(f(1+0.01) = (1.01)^2 + 3(1.01) - 7\). Calculate: \((1.01)^2 = 1.0201\), \(3 \times 1.01 = 3.03\), \(f(1.01) = 1.0201 + 3.03 - 7 = -2.9499\). Compute the difference quotient: \(\frac{f(1.01) - f(1)}{0.01} = \frac{-2.9499 - (-3)}{0.01} = 5.01\).

Calculate for h = -0.01

Now compute \(h = -0.01\) in the difference quotient.Substitute: \(f(1-0.01) = (0.99)^2 + 3(0.99) - 7\). Calculate: \((0.99)^2 = 0.9801\), \(3 \times 0.99 = 2.97\), \(f(0.99) = 0.9801 + 2.97 - 7 = -3.0499\). Compute the difference quotient: \(\frac{f(0.99) - f(1)}{-0.01} = \frac{-3.0499 - (-3)}{-0.01} = 4.99\).

Average approximations to estimate the limit

The approximated values of the derivative are \(5.1\), \(4.9\), \(5.01\), and \(4.99\). To estimate the limit of the difference quotient, average these values: \(\text{Average} = \frac{5.1 + 4.9 + 5.01 + 4.99}{4} = 5.0\). This final result suggests the limit of the difference quotient, i.e., the derivative of the function at \(a=1\), is approximately \(5.0\).

Key concepts

Difference Quotient

The difference quotient is an essential concept in calculus. It helps us understand how functions change when their inputs change by small amounts. Basically, it is a formula that helps find the slope of the secant line between two points on the graph of the function.
It is defined as:
\[ m(a+h) = \frac{f(a+h) - f(a)}{h} \]for a function \(f(x)\) and a specific point \(a\). Here, \(h\) is a small increment or decrement in \(x\). By selecting small values, like \(h = 0.1\) or \(h = -0.1\), you can start observing the behavior of the function around the point \(a\). This is very useful for gaining insight into what the derivative will be.

Derivative Approximation

The derivative gives us the rate at which a function is changing at any point. However, calculating the exact derivative involves understanding limits, as \(h\) approaches zero. But, by using small but non-zero values for \(h\), we can approximate the derivative.
In the exercise, values \( h = 0.1, -0.1, 0.01, \) and \(-0.01\) allow for different approximations of the derivative at the point \(a = 1\). When \( h \) is positive, it provides one view of the approaching slope; when \(h\) is negative, it gives another. By averaging these, we get a good estimate of the actual slope, which represents the derivative. The closer \(h\) gets to zero, the closer our approximation is to the true derivative, which in this case turns out to be around 5.0.

Polynomial Functions

Polynomial functions, like \(f(x) = x^2 + 3x - 7\), are expressions involving variables raised to whole number powers and their coefficients. They are simple yet very powerful in modeling a wide range of behaviors, from basic lines to complex curves.
For our exercise, the function is quadratic, meaning its highest power of \(x\) is 2. The behavior of this function is a parabola, a U-shaped curve. Calculating derivatives and using approximations can help us understand how steep the function is at a certain point like \( x = 1 \). This steepness corresponds to the slope of the tangent line to the curve at that point. Don't be worried if it seems tricky—we often rely on technology or stronger methods to do these calculations accurately.

Limits

The concept of limits is one of the foundational ideas in calculus. It helps us deal with values that get very close to each other. A limit essentially describes the behavior of a function as its inputs get extremely close to a number.
In the difference quotient, we calculate the limit as \(h\) approaches zero, which helps us transition from a secant line (connecting two points) to the tangent line (touching the curve at just one point). In our exercise, we made \(h\) very small but non-zero, to estimate the limit without using advanced calculus.
By calculating with small values and seeing the results converge (like our approximated derivative being very close for different \(h\) values), we can confirm our understanding or adjust our methods to get even better results as needed.