Question

use the four-step process to find r(x) and then find r(1), r(2), and r(3).

r(x) = 3 - 3x²

r(x) = \square

Explanation:

Step1: Find \( r(x + h) \)

Substitute \( x + h \) into \( r(x) = 3 - 3x^2 \):
\( r(x + h) = 3 - 3(x + h)^2 = 3 - 3(x^2 + 2xh + h^2) = 3 - 3x^2 - 6xh - 3h^2 \)

Step2: Find \( r(x + h) - r(x) \)

Subtract \( r(x) \) from \( r(x + h) \):
\( r(x + h) - r(x) = (3 - 3x^2 - 6xh - 3h^2) - (3 - 3x^2) = -6xh - 3h^2 \)

Step3: Find \( \frac{r(x + h) - r(x)}{h} \)

Divide the result by \( h \) ( \( h
eq 0 \) ):
\( \frac{r(x + h) - r(x)}{h} = \frac{-6xh - 3h^2}{h} = -6x - 3h \)

Step4: Find the limit as \( h \to 0 \)

Take the limit of \( \frac{r(x + h) - r(x)}{h} \) as \( h \) approaches 0:
\( r'(x) = \lim_{h \to 0} (-6x - 3h) = -6x \)

Now find \( r'(1) \), \( r'(2) \), and \( r'(3) \):

• For \( r'(1) \): Substitute \( x = 1 \) into \( r'(x) \): \( r'(1) = -6(1) = -6 \)
• For \( r'(2) \): Substitute \( x = 2 \) into \( r'(x) \): \( r'(2) = -6(2) = -12 \)
• For \( r'(3) \): Substitute \( x = 3 \) into \( r'(x) \): \( r'(3) = -6(3) = -18 \)

Answer:

\( r'(x) = -6x \)
\( r'(1) = -6 \)
\( r'(2) = -12 \)
\( r'(3) = -18 \)