Find intervals decrease and increase of a function f ( x ) = π x 2 − x arctan ⁡ x.Find intervals decrease and increase of a function. f ( x ) = π x 2 − x arctan ⁡ x f ′ ( x ) = π 2 − arctan ⁡ x − x 1 + x 2 f ″ ( x ) = − 2 ( 1 + x 2 ) 2 which is negative. What can I say from here?

Question

Find intervals decrease and increase of a function   f  (  x  )  =            π      x        2    −  x  arctan  ⁡  x.Find intervals decrease and increase of a function.  f  (  x  )  =            π      x        2    −  x  arctan  ⁡  x      f    ′    (  x  )  =      π    2    −  arctan  ⁡  x  −      x          1      +              x        2                  f    ″    (  x  )  =            −      2              (      1      +              x        2                    )        2            which is negative. What can I say from here?

Haylee Krause · Accepted Answer

Explanation:As       f    ″    (  x  )  &amp;lt;  0 so f′ is decreasing      f    ′    (  x  )  &amp;gt;      lim          x      →      ∞            f    ′    (  x  )  =  π      /    2  −  π      /    2  =  0so f′(x) is alsways   &amp;gt;  0.

Raina Gomez · Accepted Answer

Step 1Interval for which f increases:   (  −      ∞    ,      ∞    ). Interval which f decreases: nonef is strictly increasing, concave downward and   f  (  x  )  →  1 as   x  →  ∞.Step 2Background information.Since       f    ′    (  x  )  &amp;gt;  0,       f    ″    (  x  )  &amp;lt;  0, we conclude that f is strictly increasing and concave downward.Interval of increasing for   f  :  (  −      ∞    ,      ∞    ), interval of decreasing for f: noneAs   x  →  ∞ we have  f  (  x  )  =  1  −      1          3              x        2              +      1          5              x        4              −      1          7              x        6              +  O      (          x              −        7              )    .Hence       lim          x      →      ∞        f  (  x  )  =  1.

Find intervals decrease and increase of a function f(x)=(pi x)/2-x arctan x.

Answered question

Answer & Explanation

New Questions in Integral Calculus