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 &gt; − 2. For example, f ′ ( − 1 ) &gt; 0 and f ′ ( 5 ) &lt; 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 = 0Add 12 x 3 to both sides 6 x 2 = − 12 x 3 Multiply both sides by x 3 . 6 x = − 12Divide both sides by 6. x = − 12 6 = − 2

Question

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  &amp;gt;  −  2. For example,       f    ′    (  −  1  )  &amp;gt;  0 and       f    ′    (  5  )  &amp;lt;  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&#039;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        =  0Add       12          x      3       to both sides      6          x      2        =            −      12              x      3      Multiply both sides by       x    3  .  6  x  =  −  12Divide both sides by 6.  x  =            −      12        6    =  −  2

Marcel Mccullough · Accepted Answer

Step 1Notice 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 2You should therefore check for increasing/decreasing on the intervals   (  −  ∞  ,  −  2  )  ,    (  −  2  ,  0  ) and   (  0  ,  ∞  ).

Answered question

Answer & Explanation

New Questions in Integral Calculus