91Ó°ÊÓ

Function evaluation involves finding the value of a function at a specific point.
Let's start with the function given in the problem:
\( f(x) = \frac{1}{x^2} \)
To evaluate the function at a particular value of x, simply substitute the value into the function.
In this example, you're asked to evaluate the function at x = 1:
\( f(1) = \frac{1}{1^2} = 1 \)
Next, you need to find the value of the function after a small change in x.
Here, the change \( Δx \) is 0.5, so you evaluate the function at x + Δx, which is 1 + 0.5 = 1.5:
\( f(1.5) = \frac{1}{(1.5)^2} = \frac{1}{2.25} = 0.4444 \) (rounded to four decimal places).
This gives you the values of the function at two different points, which is essential for finding changes in the function value and computing derivatives.

The derivative of a function shows how the function's value changes as the input changes.
For the given function, \( f(x) = \frac{1}{x^2} \), finding its derivative involves applying the power rule.
The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
Rewrite the function as \( f(x) = x^{-2} \).
Then, apply the power rule:
\( f'(x) = -2x^{-3} = -\frac{2}{x^3} \)
Now, substitute x = 1 to find the derivative at that point:
\( f'(1) = -\frac{2}{1^3} = -2 \)
This tells you the rate at which the function changes at x = 1.
Derivatives help us understand the slope of the function and can be utilized to find approximate changes in the function's value over small intervals.

The change in the function value (\( Δy \)) represents how much the function's output shifts due to a change in the input (\( Δx \)).
To determine \( Δy \), subtract the function value at the original point from the function value at the new point:
\( Δy = f(x + Δx) - f(x) \)
For our example, this calculation becomes:
\( Δy = f(1.5) - f(1) = 0.4444 - 1 = -0.5556 \) (rounded to four decimal places).
Additionally, you can approximate the change in the y-value using the derivative.
This is done by multiplying the derivative at the original point by Δx:
\( f'(x) Δx = (-2)(0.5) = -1.00 \) (rounded to two decimal places).
By comparing \( Δy \) and \( f'(x) Δx \), you see that they both provide insight into how the function behaves over small changes.
While \( Δy \) gives you the exact change, \( f'(x) Δx \) offers an approximation, highlighting the power of derivatives in making predictions.