Is #f(x)=(-x^2-2x-2)/(x-3)# increasing or decreasing at #x=1#?

Question

Ezra Adams · Accepted Answer

increasing at x = 1

Finding the value of f'(1) is necessary to determine whether the function is increasing or decreasing. • At x = 1, f(x) is increasing if f'(1) > 0. • At x = 1, f(x) is decreasing if f'(1) < 0. Differentiate using the #color(blue)" Quotient rule " # If f(x) # = g(x)/(h(x)) " then " f'(x) = (h(x)g'(x) - g(x)h'(x))/[h(x)]^2# hence #f'(x) = ((x-3) d/dx(x^2-2x-2) - (x^2-2x-2) d/dx(x-3))/(x-3)^2# #= ((x-3)(2x-2) - (x^2-2x-2).1)/(x-3)^2# #f'(1) = ((1-3)(2-2) - (1-2-2))/(1-3)^2 = (0-(-3))/4 = 3/4 # #rArr " since " f'(1) > 0 " then f(x) is increasing at x = 1 "# This can be seen on graph of function. graph{(-x^2-2x-2)/(x-3) [-10, 10, -5, 5]}

Evelyn Agard · Answer

undefined To determine whether $ f(x) = \frac{-x^2 - 2x - 2}{x - 3} $ is increasing or decreasing at $ x = 1 $, we can analyze the sign of the derivative of $ f(x) $ at that point.

Taking the derivative of $ f(x) $ using the quotient rule, we get:

$$ f'(x) = \frac{(x - 3)(-2x - 2) - (-x^2 - 2x - 2)(1)}{(x - 3)^2} $$

Simplify the expression:

$$ f'(x) = \frac{-2x^2 - 2x - 6 + 2x^2 + 4x + 2}{(x - 3)^2} $$
$$ f'(x) = \frac{2x + 4}{(x - 3)^2} $$

Now, plug in $ x = 1 $:

$$ f'(1) = \frac{2(1) + 4}{(1 - 3)^2} $$
$$ f'(1) = \frac{6}{4} $$
$$ f'(1) = \frac{3}{2} $$

Since the derivative is positive at $ x = 1 $, $ f(x) $ is increasing at that point.