Is f(x) increasing or decreasing? on ( − 1 , 0 ]Let the function f(x) be defined by f ( x ) = ∑ n = 0 ∞ ( − 1 ) n − 1 ( 2 n + 1 ) ! x n . Since f(x) is a function, then it can increase and decrease. So, in the interval ( − 1 , 0 ], is f(x) increasing or decreasing? Here we restrict x to any real number between this interval. Fortunately, I found from a previous question I asked that when x &lt; 0, f ( x ) = sinh ⁡ ( x ) x . Credits to the people that answered that question. Graphing this function, I found that it is decreasing throughout. Of course, at x = 0, the entire function is 0, and it is neither decreasing or increasing.

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Is f(x) increasing or decreasing? on   (  −  1  ,  0  ]Let the function f(x) be defined by   f  (  x  )  =      ∑          n      =      0              ∞                  (      −      1              )                  n          −          1                            (      2      n      +      1      )      !            x    n  . Since f(x) is a function, then it can increase and decrease. So, in the interval   (  −  1  ,  0  ], is f(x) increasing or decreasing? Here we restrict x to any real number between this interval. Fortunately, I found from a previous question I asked that when   x  &amp;lt;  0,   f  (  x  )  =            sinh      ⁡      (              x            )              x      . Credits to the people that answered that question. Graphing this function, I found that it is decreasing throughout. Of course, at   x  =  0, the entire function is 0, and it is neither decreasing or increasing.

Cassie Moody · Accepted Answer

Step 1You don&#039;t even need to find a closed form for f. If   x  ∈  (  −  1  ,  0  ], then the series converges, and       |    x      |    =  −  x. Step 2We therefore have      f    ′    (  x  )  =      ∑          n      =      0        ∞              n      (      −      1              )                  n          −          1                            (      2      n      +      1      )      !            x          n      −      1        =      ∑          n      =      0        ∞              n      (      −      1              )                  n          −          1                            (      2      n      +      1      )      !        (  −  1      )          n      −      1            |    x            |              n      −      1        =      ∑          n      =      0        ∞        n          (      2      n      +      1      )      !            |    x            |              n      −      1        &amp;gt;  0  ,, for all   x  ∈  (  −  1  ,  0  ] except   x  =  0. From this we can conclude via the MVT that f is strictly increasing on this interval.

Is f(x) increasing or decreasing? on (-1,0]

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