Function f(x)=6+6/x+6/x^2 and I want to find the intervals where it increases or decreases. The problem is that when I find f′(x)=0, which becomes x=-2. Once I put -2 on a number line, and find whether the numbers higher and lower than -2 produce positive or negative numbers when plugged into f′(x), I get both positives and negatives for different numbers when x>-2. For example, f′(-1)>0 and f′(5)<0.

Jensen Mclean 2022-10-02 Answered
Why do I get different results when testing for increasing/decreasing intervals of a function here?
I have the function f ( x ) = 6 + 6 x + 6 x 2 and I want to find the intervals where it increases or decreases. The problem is that when I find f ( x ) = 0, which becomes x = 2. Once I put -2 on a number line, and find whether the numbers higher and lower than -2 produce positive or negative numbers when plugged into f′(x), I get both positives and negatives for different numbers when x > 2. For example, f ( 1 ) > 0 and f ( 5 ) < 0.
I am not sure if this is because the function has a vertical asymptote at x = 0 and a horizontal attribute at y = 6. If so, I'd like to know what I need to do to handle them.
How I found the first derivative:
d d x 6 x = [ 0 ] [ 6 1 ] x 2 = 6 x 2
d d x 6 x 2 = [ 0 ] [ 6 2 x ] ( x 2 ) 2 = 12 x 3
f ( x ) = 6 x 2 12 x 3
How I found the asymptotes:
When you combine the fractions of f(x), f ( x ) = 6 x 2 + 6 x + 6 x 2 . When set equal to zero, x 2 has an x-value of zero. Therefore, f(x) has a vertical asymptote at x = 0.
Once the fractions are combined, and when everything but the coefficients of the leading terms on the numerator and denominator are removed, f ( x ) = 6 1 . So f(x) has a horizontal asymptote at y = 6.
How I found the value of f ( x ) = 0:
6 x 2 12 x 3 = 0
Add 12 x 3 to both sides
6 x 2 = 12 x 3
Multiply both sides by x 3 .
6 x = 12
Divide both sides by 6.
x = 12 6 = 2
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Answers (1)

Marcel Mccullough
Answered 2022-10-03 Author has 11 answers
Step 1
Notice that f ( x ) = 6 x 2 12 x 3 = 6 x 3 ( x + 2 ).
So f′ is undefined at x = 0 and f has slope 0 at x = 2.
Step 2
You should therefore check for increasing/decreasing on the intervals ( , 2 ) , ( 2 , 0 ) and ( 0 , ).
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