91Ó°ÊÓ

In differential calculus, \(\text{\Delta y}\) represents the change in the value of a function as the input value changes. It shows how much the output increases or decreases when the input is incremented by a small amount.

For the function y = f(x) = \(\frac{1}{x}\), we can calculate the change in y (\text{\Delta y}) by evaluating the function at the original point (x) and at the incremented point (x + \(\text{\Delta x}\)).

Using the given values, x = 1 and \(\text{\Delta x}\) = 0.2, we compute:
\[ \Delta y = \frac{1}{1 + 0.2} - \frac{1}{1} = \frac{1}{1.2} - 1 \]
Simplifying the equation, we get:
\[ \Delta y = 0.8333 - 1 = -0.1667 \]
So, the change in y is -0.1667 when rounded to four decimal places. This means the output value decreases by 0.1667 units when the input value is increased by 0.2 units.

The derivative of a function, represented as f'(x), is a key concept in differential calculus. It represents the rate at which the function value changes as its input changes.

For our function y = f(x) = \(\frac{1}{x}\), the derivative f'(x) can be calculated using rules of differentiation. The derivative of \(\frac{1}{x}\) is:
\[ f'(x) = -\frac{1}{x^2} \]
The negative sign indicates that the function is decreasing as x increases. This result will help us understand the change in the function's value relative to changes in x.

Function evaluation is the process of finding the value of a function for a specific input. To evaluate a function, we substitute the input value into the function's formula.

In this exercise, we evaluate the function y = f(x) = \(\frac{1}{x}\) at different points. First, we evaluate f(x) at x = 1, giving us:
\[ f(1) = \frac{1}{1} = 1 \]
Next, we evaluate f(x) at x + \(\text{\Delta x}\), where \(\text{\Delta x}\) = 0.2:
\[ f(1 + 0.2) = f(1.2) = \frac{1}{1.2} = 0.8333 \]
Through these evaluations, we see how the function's output changes with different inputs. This is a fundamental step in finding the change in y (\text{\Delta y}) and understanding the function's behavior.

Increment calculation involves determining the small change applied to the input value of a function. This small change is denoted as \(\text{\Delta x}\).

We know that \(\text{\Delta x}\) is the increment applied to x, which means we are evaluating the function at a point slightly beyond x. This calculation helps to approximate the change in the function's output.

To find the product of the derivative and the increment, we use:
\[ f'(x) \Delta x = -\frac{1}{x^2} \Delta x = -\frac{1}{1^2} \Delta x = -1 \times 0.2 = -0.2 \]
By substituting x = 1 and \(\text{\Delta x}\) = 0.2, we find that the product results in -0.2 (rounded to two decimal places). This calculation informs us about the expected change in the function's value due to a small increment in x.