Question 1014593
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Everything you did answer is spot-on correct.


<b>Part e</b>


Intervals of increase/decrease.  Intuitively, an interval of increase is any interval where the function gets bigger as you go from left to right.  An interval of decrease is an interval where the function gets smaller as you move from left to right.  Precisely speaking, A function is increasing on an interval if, and only if, for all pairs of numbers *[tex \Large x_1\ <\ x_2] in the interval, *[tex \Large f(x_1)\ <\ f(x_2)]. The way to find the intervals is to find the intervals where the first derivative is positive and the intervals where the first derivative is negative.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}}{\text{d}x}2e^{-5x^2}\ =\ -20xe^{-5x^2}]


So we need the solution interval for:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -20xe^{-5x^2}\ >\ 0]


But since *[tex \Large e^{-5x^2}\ >\ 0] for the entire domain, we only have to solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -20x\ >\ 0]


Which, as a quick examination of the graph would suggest, is the interval *[tex \Large \left(-\infty, 0\right)].


Similarly, your function decreases on *[tex \Large \left(0, \infty)]


In point of fact, since the terms "increasing" and "decreasing" are only defined on an interval, you can't actually exclude a single point in the definition of that interval.  See discussion:  <a href="http://teachingcalculus.com/2012/11/02/open-or-closed/"><u>Open or Closed?</u> by Lin McMullin</a>



Based on that, your intervals should be:  Increasing on *[tex \Large \left(-\infty, 0\right\]] and Decreasing on [*[tex \Large \left\.0, \infty\right)]


<b>Part g</b>


Concavity is determined by the sign of the second derivative.


So you need to solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 200x^2\ -\ 20\ \geq\ 0]


for the two concave up intervals, and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 200x^2\ -\ 20\ \leq\ 0]


for the concave down interval.


As you might suspect, the critical points are the two inflection points that you found in Part h.


Here is the graph to compare with your effort:


*[illustration IncreasingDecreasingConcavityIntervals.jpg]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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