Graphing using derivativesSketch the graph of the following equation. Show steps of finding out critical numbers, intervals of increase and decrease, absolute maximum and minimum values and concavity. y = x e x 2 I found the first derivative which is y ′ = ( 2 x 2 + 1 ) e x 2 and I know that in order to find min and max the zeroes for y′ must be found, but y′ doesn't have any real zeroes, and I'm confused about how to go on with solving the problem.If someone could help me out, that would be appreciated. Thank you in advance.

Question

Graphing using derivativesSketch the graph of the following equation. Show steps of finding out critical numbers, intervals of increase and decrease, absolute maximum and minimum values and concavity.  y  =  x      e                  x        2            I found the first derivative which is       y    ′    =  (  2      x    2    +  1  )      e                  x        2             and I know that in order to find min and max the zeroes for y′ must be found, but y′ doesn&#039;t have any real zeroes, and I&#039;m confused about how to go on with solving the problem.If someone could help me out, that would be appreciated. Thank you in advance.

Bryant Liu · Accepted Answer

Step 1If   f  (  x  )  =  x      e                  x        2            , then       f    ′    (  x  )  =  2      x    2        e                  x        2              +      e                  x        2              =      e                  x        2              (  2      x    2    +  1  ), as you have written.Since the derivative never equals zero or is undefined across the domain, you can conclude there are no critical numbers and no relative extrema.Step 2Can you show that the function is always increasing by showing that       f    ′    (  x  )  &amp;gt;  0 for all x?To determine the concavity of the function, you must now find f′′(x). Can you proceed?

Garrett Sheppard · Accepted Answer

Step 1You have discovered that the function is increasing, because its derivative is positive, so the function has no maximum nor minimum.Compute also       lim          x      →      −      ∞        x      e                  x        2              =  −  ∞  ,        lim          x      →      ∞        x      e                  x        2              =  ∞  , and note there&#039;s no oblique asymptote.Step 2For concavity and convexity, compute the second derivative       f    ″    (  x  )  =  4  x      e                  x        2              +  (  2      x    2    +  1  )  ⋅  2  x      e                  x        2              =  2  (  2      x    3    +  3  x  )      e                  x        2            .The second derivative only vanishes at 0.

Sketch the graph of the following equation. Show steps of finding out critical numbers, intervals of increase and decrease, absolute maximum and minimum values and concavity. y=xe^{x^2}

Answered question

Answer & Explanation

New Questions in Integral Calculus