Question

section 2.7 derivatives and rates of change

use the definition of the derivative to find the derivative of each function with respect to \\(x\\).

1. \\(y = -2x + 5\\)

at \\(x = 4\\)

1. \\(f(x) = -4x - 2\\)

at \\(x = 5\\)

Explanation:

State the definition of the derivative

The derivative of a function \(f(x)\) at a point \(x = a\) is defined by the limit:

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

Solve Question 1

For \(y = f(x) = -2x + 5\) at \(x = 4\):

$$f(4) = -2(4) + 5 = -3$$

$$f(4+h) = -2(4+h) + 5 = -8 - 2h + 5 = -3 - 2h$$

Substitute these values into the limit definition:

$$f'(4) = \lim_{h \to 0} \frac{(-3 - 2h) - (-3)}{h} = \lim_{h \to 0} \frac{-2h}{h} = -2$$

Solve Question 2

For \(f(x) = -4x - 2\) at \(x = 5\):

$$f(5) = -4(5) - 2 = -22$$

$$f(5+h) = -4(5+h) - 2 = -20 - 4h - 2 = -22 - 4h$$

Substitute these values into the limit definition:

$$f'(5) = \lim_{h \to 0} \frac{(-22 - 4h) - (-22)}{h} = \lim_{h \to 0} \frac{-4h}{h} = -4$$

Answer:

Question 1

The derivative of \(y = -2x + 5\) at \(x = 4\) is \(-2\).

Question 2

The derivative of \(f(x) = -4x - 2\) at \(x = 5\) is \(-4\).