Question

For each function, find f(1) and f(1.01) and use the results to approximate ƒ′(1).

Answer 1

To approximate ƒ′(1), find the values of f(1) and f(1.01) and calculate the average rate of change.

Step-by-step explanation:

To approximate ƒ′(1), we need to find the values of f(1) and f(1.01) and use them to calculate the average rate of change.

Let's say the given function is f(x).

Step 1: Find f(1) by substituting x = 1 into the function.

Step 2: Find f(1.01) by substituting x = 1.01 into the function.

Step 3: Use the values of f(1) and f(1.01) to calculate the average rate of change (ƒ′(1)) using the formula: ƒ′(1) ≈ (f(1.01) - f(1))/(1.01 - 1).

Answer 2

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

To approximate the derivative of a function at a point, we can use a small value of $h$ and plug it into the difference quotient. For example, if we want to approximate $f'(1)$, we can use $h = 0.01$ and calculate:

$$f'(1) \approx \frac{f(1 + 0.01) - f(1)}{0.01}$$

This formula requires us to find the values of $f(1)$ and $f(1 + 0.01)$ for each function. Here are the functions and their values:

- $f(x) = x^2 + 3x - 2$
- $f(1) = 1^2 + 3 \times 1 - 2 = 2$
- $f(1 + 0.01) = (1 + 0.01)^2 + 3 \times (1 + 0.01) - 2 \approx 2.0601$
- $f'(1) \approx \frac{2.0601 - 2}{0.01} = 6.01$
- $f(x) = \sqrt{x}$
- $f(1) = \sqrt{1} = 1$
- $f(1 + 0.01) = \sqrt{1 + 0.01} \approx 1.004987$
- $f'(1) \approx \frac{1.004987 - 1}{0.01} = 0.4987$
- $f(x) = \sin x$
- $f(1) = \sin 1 \approx 0.841471$
- $f(1 + 0.01) = \sin (1 + 0.01) \approx 0.84622$
- $f'(1) \approx \frac{0.84622 - 0.841471}{0.01} = 0.4749$

Therefore, the approximate values of the derivatives of the functions at $x = 1$ are:

- $f'(1) \approx 6.01$ for $f(x) = x^2 + 3x - 2$
- $f'(1) \approx 0.4987$ for $f(x) = \sqrt{x}$
- $f'(1) \approx 0.4749$ for $f(x) = \sin x$