Question

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

f(x) = -x² + 6x - 3

f(x) = □

Explanation:

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

Substitute \( x + h \) into \( f(x) \):
\( f(x + h) = -(x + h)^2 + 6(x + h) - 3 \)
Expand \( (x + h)^2 \): \( (x + h)^2 = x^2 + 2xh + h^2 \)
So, \( f(x + h) = -(x^2 + 2xh + h^2) + 6x + 6h - 3 = -x^2 - 2xh - h^2 + 6x + 6h - 3 \)

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

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

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

Divide the result from Step 2 by \( h \) ( \( h
eq 0 \)):
\( \frac{f(x + h) - f(x)}{h} = \frac{-2xh - h^2 + 6h}{h} = \frac{h(-2x - h + 6)}{h} = -2x - h + 6 \) (Cancel \( h \))

Step4: Find \( f'(x) \) by taking the limit as \( h \to 0 \)

\( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} (-2x - h + 6) \)
As \( h \to 0 \), \( -h \to 0 \), so \( f'(x) = -2x + 6 \)

Answer:

\( f'(x) = -2x + 6 \)