Is #f(x)=(6x^2-x-12)/(x+3)# increasing or decreasing at #x=3#?

Question

Charles Anctil · Accepted Answer

#"increasing at " x=3# #"to determine if f(x) is increasing/decreasing at x = a"# #"differentiate and evaluate at x = a"# #• " if "f'(a)>0" then f(x) is increasing at x = a"# #• " if " f'(a)<0" then f(x) is decreasing at x = a"# #"differentiate f(x) using the "color(blue)"quotient rule"# #"given " f(x)=(g(x))/(h(x))" then"# #f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2# #g(x)=6x^2-x-12rArrg'(x)=12x-1# #h(x)=x+3rArrh'(x)=1# #rArrf'(x)=((x+3)(12x-1)-(6x^2-x-12))/(x+3)^2# #rArrf'(3)=(210-39)/36>0# #rArrf(x)" is increasing at "x=3#

William Abdalla · Answer

undefined To determine if the function $ f(x) = \frac{6x^2 - x - 12}{x + 3} $ is increasing or decreasing at $ x = 3 $, we can use the first derivative test.

1. Find the first derivative of $ f(x) $.
2. Evaluate the first derivative at $ x = 3 $.
3. If the first derivative is positive, the function is increasing at $ x = 3 $. If it's negative, the function is decreasing at $ x = 3 $.

The first derivative of $ f(x) $ is $ f'(x) = \frac{18x^2 + 12x - 15}{(x + 3)^2} $.

Evaluate $ f'(3) $:

$$ f'(3) = \frac{18(3)^2 + 12(3) - 15}{(3 + 3)^2} $$
$$ f'(3) = \frac{18(9) + 12(3) - 15}{36} $$
$$ f'(3) = \frac{162 + 36 - 15}{36} $$
$$ f'(3) = \frac{183}{36} $$

Since $ f'(3) > 0 $, the function $ f(x) $ is increasing at $ x = 3 $.